Two Dimensional Geometry jee mains questions | JEE Main Two Dimensional Geometry Previous Year Question

Q 1 :

Let the line 2x + 3y – k = 0, k > 0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line segment AB as a diameter is $x^{2} + y^{2} - 3 x - 2 y = 0$ and the length of the latus rectum of the ellipse $x^{2} + 9 y^{2} = k^{2}$ is $\frac{m}{n}$ , where m and n are coprime, then 2m + n is equal to [2024]

10
13
12
11

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(4)

Equation of circle with AB as diameter is

$(x - \frac{k}{2}) x + y (y - \frac{k}{3}) = 0$

$\Rightarrow x^{2} + y^{2} - \frac{k x}{2} - \frac{k y}{3} = 0$

On comparing it with given equation $x^{2} + y^{2} - 3 x - 2 y = 0$

we get $\frac{k}{2} = 3 \Rightarrow k = 6$

Now, equation of ellipse is given by

$x^{2} + 9 y^{2} = 36$

$\Rightarrow \frac{x^{2}}{36} + \frac{y^{2}}{4} = 1$

So latus rectum $\frac{2 b^{2}}{a} = \frac{2 \times 4}{6} = \frac{4}{3} = \frac{m}{n}$

$\Rightarrow 2 m + n = 2 \times 4 + 3 = 11$ .

Q 2 :

Let $f (x) = x^{2} + 9$ , $g (x) = \frac{x}{x - 9}$ and a = fog (10), b = gof (3). If e and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{a} + \frac{y^{2}}{b} = 1$ , then $8 e^{2}$ + $l^{2}$ is equal to [2024]

6
12
16
8

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(4)

We have, $g (x) = \frac{x}{x - 9} so, g (10) = 10$

Also $f (x) = x^{2} + 9 so, f (10) = 10^{2} + 9 = 109$

$a = f o g (10) = f (g (10)) = f (10) = 109$

Also, $f (3) = 3^{2} + 9 = 18$

Now, $b = g o f (3) = g (f (3)) = g (18) = \frac{18}{18 - 9} = 2$

So ellipse $\frac{x^{2}}{a} + \frac{y^{2}}{b} = 1$ becomes $\frac{x^{2}}{109} + \frac{y^{2}}{2} = 1$

$∴ e = \sqrt{1 - \frac{2}{109}} = \sqrt{\frac{107}{109}}$

and $l$ $= \frac{2 b^{2}}{a} = \frac{2 \times 2}{\sqrt{109}} = \frac{4}{\sqrt{109}}$

Hence, $8 e^{2}$ + $l^{2}$ $= \frac{8 \times 107}{109} + \frac{16}{109} = 8$ .

Q 3 :

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum is $\sqrt{14}$ . Then the square of the eccentricity of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is: [2024]

3
3/2
7/2
5/2

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(2)

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$

$e = \frac{1}{\sqrt{2}} \Rightarrow \sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{1}{\sqrt{2}} \Rightarrow 1 - \frac{b^{2}}{a^{2}} = \frac{1}{2} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{1}{2}$

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 : e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{1}{2} = \frac{3}{2}$

Q 4 :

Let P be a point on the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ , Let the line passing through P and parallel to y-axis meet the circle $x^{2} + y^{2} = 9$ at point Q such that P and Q are on the same side of the x-axis. then, the eccentricity of the locus of the point R and PQ such that PR : RQ = 4 : 3 as P moves on the ellipse, is [2024]

$\frac{13}{21}$
$\frac{\sqrt{139}}{23}$
$\frac{\sqrt{13}}{7}$
$\frac{11}{19}$

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(3)

As, P be a point on the ellipse.

$∴$ Coordinates of P = (3 cos $θ$ , 2 sin $θ$ )

Also, Q is a point on the circle $x^{2} + y^{2} = 9$

$∴$ Coordinates of Q = (3 cos $θ$ , 3 sin $θ$ )

Now, R(h, k) divides the line PQ in the ratio 4 : 3.

$∴ h = \frac{12 \cos θ + 9 \cos θ}{7} = \frac{21 \cos θ}{7} = 3 \cos θ$

and $k = \frac{12 \sin θ + 6 \sin θ}{7} = \frac{18}{7} \sin θ$

Now, $\frac{h^{2}}{9} + \frac{49 k^{2}}{324} = 1$

So, locus of R, $\frac{x^{2}}{3^{2}} + \frac{y^{2}}{{(18 / 7)}^{2}} = 1$ is ellipse.

$∴$ Eccentricity = $\sqrt{1 - \frac{{(18 / 7)}^{2}}{9}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7}$ .

Q 5 :

The length of the chord of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid point is $(1, \frac{2}{5})$ , is equal to : [2024]

$\frac{\sqrt{1541}}{5}$
$\frac{\sqrt{1691}}{5}$
$\frac{\sqrt{1741}}{5}$
$\frac{\sqrt{2009}}{5}$

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(2)

Ellipse : $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ ... (i)

Equation of chord having mid point $(x_{1}, y_{1})$ is

$\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}}$

$\Rightarrow \frac{x}{25} + \frac{y}{16} (\frac{2}{5}) = \frac{1}{25} + \frac{1}{16} (\frac{4}{25}) \Rightarrow \frac{x}{25} + \frac{y}{40} = \frac{1}{25} + \frac{1}{100}$

$\Rightarrow 8 x + 5 y = 10$

$\Rightarrow 8 x = 5 (2 - y)$ ...(ii)

From equation (i) & (ii), we get

$x = 1 \pm \frac{\sqrt{19}}{2}, y = \frac{2}{5} (1 \mp 2 \sqrt{19)}$

Length of Chord

$= \sqrt{{[1 + \frac{\sqrt{19}}{2} - (1 - \frac{\sqrt{19}}{2})]}^{2} + {[\frac{2}{5} (1 - 2 \sqrt{19}) - \frac{2}{5} (1 + 2 \sqrt{19})]}^{2}}$

$= \sqrt{19 + \frac{64}{25} \times 19} = \frac{\sqrt{1691}}{5}$

Q 6 :

If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is : [2024]

$\frac{\sqrt{5}}{3}$
$\frac{2}{\sqrt{5}}$
$\frac{\sqrt{3}}{2}$
$\frac{1}{\sqrt{3}}$

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(2)

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be the given ellipse

We have, minor axis $\frac{1}{2} (2 a e)$ , where e is eccentricity

$\Rightarrow 2 b = a e \Rightarrow 4 b^{2} = a^{2} e^{2}$

$\Rightarrow 4 b^{2} = a^{2} - b^{2}$ [ $∵ b^{2} = a^{2} - a^{2} e^{2}$ ]

$\Rightarrow 5 b^{2} = a^{2}$

Now, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$

Q 7 :

Let A( $α$ , 0) and B(0, $β$ ) be the points on the line 5x + 7y = 50. Let the point P divide the line segment AB internally in the ratio 7 : 3. Let 3x – 25 = 0 be a directrix of the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , and the corresponding focus be S. If from S, the perpendicular on the x-axis passes through P, then the length of the latus rectum of E is equal to, [2024]

$\frac{32}{5}$
$\frac{32}{9}$
$\frac{25}{9}$
$\frac{25}{3}$

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(1)

A( $α$ , 0) and B(0, $β$ ) be the points on the line 5x + 7y = 50

$\Rightarrow α = 10 and β = \frac{50}{7}$

Using section formula, we have P(3, 5).

Directrix : $x = \frac{25}{3} = \frac{a}{e}$

The equation of the line passing through P(3, 5) and perpendicular to x-axis is x = 3.

$∵$ The perpendicular is also passes through S.

$∴$ ae = 3

$a \times \frac{3}{25} a = 3 or a^{2} = 25 \Rightarrow a = 5$

$b^{2} = 25 (1 - \frac{9}{25}) = 16$

$∴$ Length of latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 16}{5} = \frac{32}{5}$ .

Q 8 :

Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6. Let an ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola P. Then the square of the length of the latus rectum of E is [2024]

$\frac{385}{8}$
$\frac{656}{25}$
$\frac{347}{8}$
$\frac{512}{25}$

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(2)

Given that vertex of parabola P is (2, 3) and equation of directrix is 2x + y = 6 ... (i)

Let S( $α$ , $β$ ) be the focus of parabola.

Equation of axis of AS is

$y - 3 = \frac{1}{2} (x - 2)$

$\Rightarrow 2 y - 6 = x - 2$

$\Rightarrow x - 2 y + 4 = 0$ ... (ii)

Solving equations (i) and (ii), we get

$∴ x = \frac{8}{5} and y = \frac{14}{5} ∴ A \equiv (\frac{8}{5}, \frac{14}{5})$

Now, using mid point formula, we have

$\frac{\frac{8}{5} + α}{2} = 2; \frac{\frac{14}{5} + β}{2} = 3 \Rightarrow \frac{8}{5} + α = 4; \frac{14}{5} + β = 6$

$\Rightarrow α = 4 - \frac{8}{5} and β = 6 - \frac{14}{5} \Rightarrow α = \frac{12}{5}; and β = \frac{16}{5}$

The point $S (\frac{12}{5}, \frac{16}{5})$ will satisfy the ellipse.

$∴ \frac{144}{25 a^{2}} + \frac{256}{25 b^{2}} = 1$ ... (iii)

Now, $b^{2} = a^{2} (1 - e^{2})$

$b^{2} = a^{2} (1 - \frac{1}{2})$

$∴$ From (iii), $\frac{144}{25 \times 2 b^{2}} + \frac{256}{25 b^{2}} = 1$

$\Rightarrow \frac{72}{25} + \frac{256}{25} = b^{2} \Rightarrow b^{2} = \frac{328}{25} ∴ a^{2} = \frac{2 \times 328}{25} = \frac{656}{25}$

$∴$ Length of latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 328}{25 \times \frac{\sqrt{656}}{5}} = \frac{656 \times 5}{25 \times \sqrt{656}} = \frac{\sqrt{656}}{5}$

$∴$ Square of the latus rectum $= \frac{656}{25}$ .

Q 9 :

If the points of intersection of two distinct conics $x^{2} + y^{2} = 4 b$ and $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ lie on the curve $y^{2} = 3 x^{2}$ , then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is _________. [2024]

(432)

We have, $x^{2} + y^{2} = 4 b$ ... (i)

and $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ ... (ii)

From (i) and (ii), $x^{2} = \frac{16 b}{b + 4}$ and $y^{2} = \frac{4 b^{2}}{b + 4}$

$∵$ These points lie on curve $y^{2} = 3 x^{2}$ .

So, $\frac{4 b^{2}}{b + 4} = 3 \times \frac{16 b}{b + 4}$

$\Rightarrow b^{2} = 12 b \Rightarrow b (b - 12) = 0$

$\Rightarrow b = 12 (b \neq 0)$

So, the points are $(\pm 2 \sqrt{3}, \pm 6)$

$∴$ Area of rectangle = $4 \sqrt{3} \times 12 = 48 \sqrt{3}$ sq. units

Thus, $3 \sqrt{3}$ times of the area of rectangle $= 3 \sqrt{3} \times 48 \sqrt{3} = 432$ .

Q 10 :

If S and $S'$ are the foci of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{9} = 1$ and P be a point on the ellipse, then $\min (S P \cdot S' P) + \max (S P \cdot S' P)$ is equal to : [2025]

27
$3 (6 + \sqrt{2})$
9
$3 (1 + \sqrt{2})$

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(1)

We have, $\frac{x^{2}}{18} + \frac{y^{2}}{9} = 1$

$\Rightarrow a = 3 \sqrt{2} and b = 3$

Since, $P S + P S' = 2 a = 2 \times 3 \sqrt{2} = 6 \sqrt{2}$

Also, $b^{2} = a^{2} (1 - e^{2}) \Rightarrow 9 = 18 (1 - e^{2})$

$\Rightarrow e = \frac{1}{\sqrt{2}}$

Equation of Directrix is $X = \frac{a}{e} = \frac{3 \sqrt{2}}{\frac{1}{\sqrt{2}}} = 6$

$S P \cdot S P' =$ $| \frac{1}{\sqrt{2}} (3 \sqrt{2} \cos θ - 6) \cdot \frac{1}{\sqrt{2}} (3 \sqrt{2} \cos θ + 6) |$

$= \frac{1}{2}$ $| 18 \cos^{2} θ - 36 |$ $=$ $| 9 \cos^{2} θ - 18 |$

${(S P \cdot S P')}_{\max} = 18; {(S P \cdot S P')}_{\min} = 9$ [ $∵ \cos^{2} x \in [0, 1]$ ]

$∴$ Required sum = 27.

Q 11 :

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is : [2025]

$\frac{3}{\sqrt{19}}$
$\frac{\sqrt{5}}{7}$
$\frac{\sqrt{3}}{16}$
$\frac{4}{\sqrt{17}}$

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(4)

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be the given ellipse with length of minor axis as 2b and distance between foci is 2ae.

$∴ 2 b = \frac{1}{4} \times (2 a e) \Rightarrow 4 b = a e \Rightarrow \frac{b}{a} = \frac{e}{4}$ ... (i)

We know, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{e^{2}}{16}}$ [From (i)]

$\Rightarrow e^{2} = \frac{16 - e^{2}}{16} \Rightarrow 16 e^{2} = 16 - e^{2} \Rightarrow 17 e^{2} = 16$

$e = \frac{4}{\sqrt{17}}$ .

Q 12 :

A line passing through the point $P (\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$ at A and B such that (PA) $\cdot$ (PB) is maximum. Then $5 (P A^{2} + P B^{2})$ is equal to : [2025]

338
377
218
290

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(1)

$\frac{x - \sqrt{5}}{\cos θ} = \frac{y - \sqrt{5}}{\sin θ} = r$

$x = r \cos θ + \sqrt{5}$

$y = r \sin θ + \sqrt{5}$

$25 x^{2} + 36 y^{2} = 900, (x, y)$ lie on E

$r^{2} [25 \cos^{2} θ + 36 \sin^{2} θ] + r [50 \sqrt{5} \cos θ + 72 \sqrt{5} \sin θ] + 25 [5] + 36 [5] = 900$

$r_{1} r_{2} = \frac{900 - 305}{25 + 11 \sin^{2} θ}, {(r_{1} r_{2})}_{\max} = \frac{595}{25} = \frac{119}{5}$

${| r_{1} \cdot r_{2} |}_{\max} = (\frac{595}{25}) = (\frac{119}{5}) at θ = 0$

$\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$

At $y = \sqrt{5}$

$\frac{x^{2}}{36} + \frac{1}{5} = 1 \Rightarrow \frac{x^{2}}{36} = \frac{4}{5} \Rightarrow x = \pm \frac{12}{\sqrt{5}}$

$P A^{2} + P B^{2} = {(P A + P B)}^{2} - 2 P A \cdot P B = {(\frac{24}{\sqrt{5}})}^{2} - 2 \cdot \frac{119}{5}$

$5 (P A^{2} + P B^{2}) = 5 (\frac{24^{2}}{5} - \frac{2.119}{5}) = 24^{2} - 238 = 338$ .

Q 13 :

Let C be the circle of minimum area enclosing the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with eccentricity $\frac{1}{2}$ and foci $(\pm 2, 0)$ . Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 2a is parallel to the major axis of E and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is : [2025]

$6 (2 + \sqrt{3})$
$8 (3 + \sqrt{2})$
$6 (3 + \sqrt{2})$
$8 (2 + \sqrt{3})$

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(4)

Given, foci $= (\pm 2, 0) = (\pm a e, 0)$ and eccentricity $(e) = \frac{1}{2}$

$∴ a e = 2$

$\Rightarrow a \times \frac{1}{2} = 2 \Rightarrow a = 4$ $[∵ e = \frac{1}{2}]$

Now, $b^{2} = a^{2} (1 - e^{2})$

$= 16 (1 - \frac{1}{4}) = 12 \Rightarrow b = 2 \sqrt{3}$

Since, the circle of minimum area enclosing the ellipse has a radius equal to semi-major axis i.e., a

$∴$ Radius = a = 4

Height of $△ P Q R = radius + 2 \sqrt{3} = 4 + 2 \sqrt{3}$

Hence, required area

$= \frac{1}{2} \times base \times height = \frac{1}{2} \times 2 a \times (4 + 2 \sqrt{3})$

$= 4 (4 + 2 \sqrt{3}) = 8 (2 + \sqrt{3}) sq . units$ .

Q 14 :

The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, –3) and eccentricity is $\frac{4}{5}$ , is [2025]

$\frac{6}{5}$
$\frac{10}{3}$
$\frac{18}{5}$
$\frac{50}{3}$

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(3)

We have, two foci (2, 5), (2, –3) and eccentricity =4/5

$\Rightarrow 2 b e = 8 \Rightarrow b e = 4 \Rightarrow b (\frac{4}{5}) = 4 \Rightarrow b = 5$

$∴ c^{2} = b^{2} - a^{2} \Rightarrow 16 = 25 - a^{2} \Rightarrow a = 3$

$∴$ Length of latus rectum = $\frac{2 a^{2}}{b} = \frac{18}{5}$

Q 15 :

The centre of a circle C is at the centre of the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b. Let C pass through the foci $F_{1}$ and $F_{2}$ of E such that the circle C and the ellipse E intersect at four points. Let P be one of these four points. If the area of the triangle $P F_{1} F_{2}$ is 30 and the length of the major axis of E is 17, then the distance between the foci of E is : [2025]

26
13
12
$\frac{13}{2}$

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(2)

We have, the ellipse $(E) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , (a > b)

$\Rightarrow x^{2} + \frac{a^{2} y^{2}}{b^{2}} = a^{2}$ ... (i)

The circle is $x^{2} + y^{2} = a^{2} e^{2}$ ... (ii)

Using (i) and (ii),

$\Rightarrow y^{2} (1 - \frac{a^{2}}{b^{2}}) = a^{2} (e^{2} - 1) = a^{2} (1 - \frac{b^{2}}{a^{2}} - 1) = - b^{2}$

$\Rightarrow y^{2} \frac{(b^{2} - a^{2})}{b^{2}} = - b^{2} \Rightarrow y^{2} \frac{b^{4}}{a^{2} - b^{2}} \Rightarrow | y | = \frac{b^{2}}{\sqrt{a^{2} - b^{2}}}$

Area of triangle $= \frac{1}{2} \times 2 a e \times \frac{b^{2}}{\sqrt{a^{2} - b^{2}}} = 30$

$\Rightarrow \frac{a b^{2} e}{a \sqrt{1 - \frac{b^{2}}{a^{2}}}} = b^{2} = 30$

Given, $2 a = 17 \Rightarrow a = \frac{17}{2}$ .

$∴$ Distance between the foci = 2ae

$= 17 \sqrt{1 - \frac{b^{2}}{a^{2}}} = 17 \sqrt{1 - \frac{30 \times 4}{289}} = 13$ .

Q 16 :

Let for two distinct values of p the lines y = x + p touch the ellipse $E : \frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$ at the points A and B. Let the line y = x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to : [2025]

24
36
48
20

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(1)

We have ellipse $E : \frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$

and line y = x + p $\Rightarrow$ slope, m = 1

E and line y = x + p has point of contacts as A and B.

So, the point of contact

$= (\frac{\mp a^{2} m}{\sqrt{a^{2} m^{2} + b^{2}}}, \frac{\pm b^{2}}{\sqrt{a^{2} m^{2} + b^{2}}}) = (\frac{\mp 16}{5}, \frac{\pm 9}{5})$

Then, $A (\frac{- 16}{5}, \frac{9}{5}) and B (\frac{16}{5}, \frac{- 9}{5})$

Now, line y = x intersects with ellipse E at

$D (\frac{- 12}{5}, \frac{- 12}{5}) and C (\frac{12}{5}, \frac{12}{5})$

ABCD does not form any quadrilateral but if we do not consider the order then we have,

Area of $△ A B C = \frac{1}{2}$ $| \begin{matrix} - \frac{16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & - \frac{9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1 \end{matrix} |$ $= 12$

$∴$ Area of quadrilateral ABCD = 2 (Area of $△$ ABC) = 24 sq. units

Q 17 :

Let the system of equations

x + 5y – z = 1

4x + 3y – 3z = 7

24x + y + $λ$ z = $μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

3
5
6
4

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(1)

For infinitely many solutions, we have, D = 0

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 4 & 3 & - 3 \\ 24 & 1 & λ \end{matrix} |$ $= 0$

$\Rightarrow 1 (3 λ + 3) - 5 (4 λ + 72) - 1 (4 - 72) = 0$

$\Rightarrow 17 λ = - 289 \Rightarrow λ = - 17$ ... (i)

Now, $D_{1} = 0$

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 7 & 3 & - 3 \\ μ & 1 & - 17 \end{matrix} |$ $= 0$

$\Rightarrow 1 (- 51 + 3) - 5 (- 119 + 3 μ) - 1 (7 - 3 μ) = 0$

$\Rightarrow - 48 + 595 - 15 μ - 7 + 3 μ = 0 \Rightarrow 12 μ = 540$

$\Rightarrow μ = 45$ ... (ii)

Using (i) and (ii), we get

$x + 5 y - z = 1, 4 x + 3 y - 3 z = 7, 24 x + y - 17 z = 45$

$\Rightarrow z = x + 5 y - 1$

$∴ 4 x + 3 y - 3 x - 15 y + 3 = 7$

$\Rightarrow x - 12 y = 4$

$\Rightarrow x = 4 + 12 y and z = 4 + 12 y + 5 y - 1 = 3 + 17 y$

$∴$ (x, y, z) = (4 + 12k, k, 3 + 17k) ( $∵$ Assume y = k)

Also, $7 \leq 7 + 30 k \leq 77$

$\Rightarrow 0 \leq 30 k < 70$

$\Rightarrow 0 \leq k \leq 2.3 \Rightarrow k = 0, 1, 2$

Thus, there are three possible solutions.

Q 18 :

Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{n} = 1$ is: [2025]

$\frac{\sqrt{7}}{4}$
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$\frac{3}{4}$

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(3)

Total triangles $= {}^{n}C_{3}$

Total quadrilaterals $= {}^{n}C_{4}$

$∵ {}^{n}C_{3} + {}^{n}C_{4} = 126 \Rightarrow {}^{n + 1}C_{4} = 126$

$\Rightarrow n = 8$

The ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{n} = 1 \Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{8} = 1$

$∴ e = \sqrt{1 - \frac{8}{16}} = \sqrt{\frac{8}{16}} = \frac{1}{\sqrt{2}}$

Q 19 :

Let the ellipse $3 x^{2} + p y^{2} = 4$ pass through the centre C of the circle $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ of radius r. Let $f_{1}, f_{2}$ be the focal distances of the point C on the ellipse. Then $6 f_{1} f_{2} - r$ is equal to [2025]

78
68
70
74

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(3)

Given, equation of circle is $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$

It can be written as

${(x - 1)}^{2} + {(y - 2)}^{2} = 16$

$∴$ Centre of circle is (1, 2) and radius is 4.

Now, ellipse $3 x^{2} + p y^{2} = 4$ passes through (1, 2)

$\Rightarrow 3 + 4 p = 4 \Rightarrow p = \frac{1}{4}$

$∴$ Given equation of ellipse is $3 x^{2} + \frac{1}{4} y^{2} = 4$

i.e., $\frac{x^{2}}{4 / 3} + \frac{y^{2}}{16} = 1$

$∴ a^{2} = \frac{4}{3} and b^{2} = 16 \Rightarrow a = \frac{2}{\sqrt{3}} and b = 4$

Now, eccentricity of ellipse $= \sqrt{\frac{1 - a^{2}}{b^{2}}} = \sqrt{1 - \frac{\frac{4}{3}}{16}} = \sqrt{\frac{11}{12}}$

Now, focal distance of ellipse from (1, 2) $= 4 \pm \sqrt{\frac{11}{12}} \times 2$

$∴ f_{1} = 4 + \frac{2 \sqrt{11}}{2 \sqrt{3}} = 4 + \sqrt{\frac{11}{3}} and f_{2} = 4 - \sqrt{\frac{11}{3}}$

Now, $f_{1} f_{2} = 16 - \frac{11}{3} = \frac{37}{3}$

$∴ 6 f_{1} f_{2} - r = 74 - 4 = 70$ .

Q 20 :

The length of the chord of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$ , whose mid-point is $(1, \frac{1}{2})$ , is: [2025]

$\frac{1}{3} \sqrt{15}$
$\sqrt{15}$
$\frac{2}{3} \sqrt{15}$
$\frac{5}{3} \sqrt{15}$

1

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(3)

We have, $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$ ,,, (i)

Mid-point of chord is $(1, \frac{1}{2})$

The equation of chord to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ bisected at the point $(x_{1}, y_{1})$ is given by

$\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} - 1 = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} - 1$

$∴ \frac{x \cdot 1}{4} + \frac{y \cdot 1 / 2}{2} - 1 = \frac{1}{4} + \frac{{(\frac{1}{2})}^{2}}{2} - 1$

$\Rightarrow x + y = 3 / 2$ ... (ii)

On solving equation (i) and (ii), we get

$x = \frac{6 \pm \sqrt{30}}{6} and y = \frac{3}{2} - (\frac{6 \pm \sqrt{30}}{6})$

Let $x_{1} = \frac{6 + \sqrt{30}}{6}, x_{2} = \frac{6 - \sqrt{30}}{6}$

and $y_{1} = \frac{3}{2} - (\frac{6 + \sqrt{30}}{6}), y_{2} = \frac{3}{2} - (\frac{6 - \sqrt{30}}{6})$

$∴$ The length of chord

$= \sqrt{(\frac{6 + \sqrt{30}}{6} - {(\frac{6 - \sqrt{30}}{6}))}^{2} + (\frac{3}{2} - (\frac{6 + \sqrt{30}}{6}) - \frac{3}{2} + {(\frac{6 - \sqrt{30}}{6}))}^{2}}$

$= \sqrt{\frac{30}{9} + \frac{30}{9}} = \sqrt{\frac{60}{9}}$

$= \frac{2}{3} \sqrt{15}$

Q 21 :

Let the product of the focal distances of the point $(\sqrt{3}, \frac{1}{2})$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , (a > b), be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipse is [2025]

$\frac{3 - 2 \sqrt{2}}{2 \sqrt{3}}$
$\frac{1 - 2 \sqrt{2}}{\sqrt{3}}$
$\frac{3 - 2 \sqrt{2}}{3 \sqrt{2}}$
$\frac{1 - \sqrt{3}}{\sqrt{2}}$

1

2

3

4

(1)

Product of focal distances from point $(\sqrt{3}, \frac{1}{2})$

$= (a + e x_{1}) (a - e x_{1})$

$\Rightarrow a^{2} - 3 e^{2} = \frac{7}{4}$ [ $∵ x_{1} = \sqrt{3}$ ]

$\Rightarrow 4 a^{2} = 7 + 12 e^{2}$

Also, $(\sqrt{3}, \frac{1}{2})$ lies on ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{3}{a^{2}} + \frac{1}{4 b^{2}} = 1 \Rightarrow \frac{3}{a^{2}} + \frac{1}{4 a^{2} (1 - e^{2})} = 1$ [ $∵ b^{2} = a^{2} (1 - e^{2})$ ]

$\Rightarrow 12 (1 - e^{2}) + 1 = 4 a^{2} (1 - e^{2})$

$\Rightarrow 13 - 12 e^{2} = (7 + 12 e^{2}) (1 - e^{2})$

$\Rightarrow 13 - 12 e^{2} = 7 + 12 e^{2} - 7 e^{2} - 12 e^{4}$

$\Rightarrow 12 e^{4} - 17 e^{2} + 6 = 0$

$\Rightarrow e^{2} = \frac{3}{4} or \frac{2}{3} \Rightarrow e = \frac{\sqrt{3}}{2} or \sqrt{\frac{2}{3}}$

$∴$ Required difference = $| \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} |$ $= \frac{3 - 2 \sqrt{2}}{2 \sqrt{3}}$

Q 22 :

The equation of the chord, of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid-point is (3, 1) is : [2025]

4x + 122y = 134
5x + 16y = 31
25x + 101y = 176
48x + 25y = 169

1

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3

4

(4)

Given: Ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and mid-point is (3, 1).

The equation of chord with given middle point is given by

T = $S_{1}$

$\Rightarrow \frac{3 x}{25} + \frac{1 y}{16} - 1 = \frac{9}{25} + \frac{1}{16} - 1$

$\Rightarrow \frac{3 x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16}$

$\Rightarrow 48 x + 25 y = 144 + 25$

$\Rightarrow 48 x + 25 y = 169$ .

Q 23 :

Let the ellipse $E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b and $E_{2} : \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$ , A < B have same eccentricity $\frac{1}{\sqrt{3}}$ . Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$ , and the distance between the foci of $E_{1}$ be 4. If $E_{1}$ and $E_{2}$ meet at A, B, C and D, then the area of the quadrilateral ABCD equals : [2025]

$6 \sqrt{6}$
$\frac{12 \sqrt{6}}{5}$
$\frac{24 \sqrt{6}}{5}$
$\frac{18 \sqrt{6}}{5}$

1

2

3

4

(3)

We have, $2 a e = 4 \Rightarrow 2 a (\frac{1}{\sqrt{3}}) = 4 \Rightarrow a = 2 \sqrt{3}$

$b^{2} = a^{2} (1 - e^{2}) = 12 (1 - \frac{1}{3}) = 8$

Now, $\Rightarrow \frac{2 b^{2}}{a} \cdot \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}}$

$\Rightarrow \frac{2 \cdot 8}{2 \sqrt{3}} \cdot \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}} \Rightarrow A^{2} = 2 B$

Alao, $\Rightarrow A^{2} = B^{2} (1 - e^{2}) \Rightarrow 2 B = B^{2} \cdot \frac{2}{3} \Rightarrow B = 3 \Rightarrow A^{2} = 6$

$∴ E_{1} : \frac{x^{2}}{12} + \frac{y^{2}}{8} = 1$ ... (i)

and $E_{2} : \frac{x^{2}}{6} + \frac{y^{2}}{9} = 1$ ... (ii)

Solving (i) and (ii), we get

$A (\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), B (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), C (\frac{\sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}}), D (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}})$

which form a rectangle.

$∴$ Required area $\frac{12}{\sqrt{5}} \times \frac{2 \sqrt{6}}{\sqrt{5}} = \frac{24 \sqrt{6}}{5}$

Q 24 :

If the mid-point of a chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is $(\sqrt{2}, \frac{4}{3})$ and the length of the chord is $\frac{2 \sqrt{α}}{3}$ , then $α$ is: [2025]

22
26
20
18

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2

3

4

(1)

Let AB is a chord and M is the mid-point.

If $M (\sqrt{2}, \frac{4}{3})$ then equation of AB is

$T = S_{1} \Rightarrow \frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} - 1 = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} - 1$

$\Rightarrow \frac{x \sqrt{2}}{9} + \frac{y}{4} (\frac{4}{3}) = \frac{{(\sqrt{2})}^{2}}{9} + \frac{{(\frac{4}{3})}^{2}}{4}$

$\Rightarrow \frac{\sqrt{2} x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9}$

$\Rightarrow \sqrt{2} x + 3 y = 6 \Rightarrow y = \frac{6 - \sqrt{2} x}{3}$

Putting in ellipse, we get $\frac{x^{2}}{9} + \frac{{(6 - \sqrt{2} x)}^{2}}{9 \times 4} = 1$

$\Rightarrow 4 x^{2} + 36 + 2 x^{2} - 12 \sqrt{2} x = 36$

$\Rightarrow 6 x^{2} - 12 \sqrt{2} x = 0 \Rightarrow 6 x (x - 2 \sqrt{2}) = 0$

$\Rightarrow x = 0 and x = 2 \sqrt{2}$

So, y = 2 and $y = \frac{2}{3}$

$∴$ Length of the chord = $= \sqrt{{(2 \sqrt{2} - 0)}^{2} + {(\frac{2}{3} - 2)}^{2}}$

$= \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{2}{3} \sqrt{22}$

So, $α = 22$ .

Q 25 :

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ , whose mid point is $(\frac{5}{2}, \frac{1}{2})$ , then $α + β$ is equal to : [2025]

46
58
37
72

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3

4

(2)

We have, equation of ellipse as, $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

Equation of chord with mid-point $(\frac{5}{2}, \frac{1}{2})$ is

$\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}}$

$\Rightarrow \frac{5}{2} (\frac{x}{9}) + \frac{1}{2} (\frac{y}{4}) = \frac{25}{36} + \frac{1}{16}$

$\Rightarrow \frac{5 x}{18} + \frac{y}{8} = \frac{109}{144}$

$\Rightarrow$ 40x + 18y = 109

On comparing with given equation $α x + β y = 109$ , we get

$α = 40 and β = 18$

$∴ α + β = 58$ .

Q 26 :

Let C be the circle $x^{2} + {(y - 1)}^{2} = 2$ , $E_{1}$ and $E_{2}$ be two ellipses whose centres lie at the origin and major axes lie on x-axis and y-axis respectively. Let the straight line x + y = 3 touch the curves C, $E_{1}$ and $E_{2}$ at $P (x_{1}, y_{1})$ , $Q (x_{2}, y_{2})$ and $R (x_{3}, y_{3})$ respectively. Given that P is the mid-point of the line segment QR and $P Q = \frac{2 \sqrt{2}}{3}$ , the value of $9 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})$ is equal to __________. [2025]

(46)

(a) Solving the line x + y = 3, and the circle $x^{2} + {(y - 1)}^{2} = 2$

Substitute y = 3 – x

$x^{2} + {(3 - x - 1)}^{2} = 2$

$\Rightarrow x^{2} - 2 x + 1 = 0$

$\Rightarrow x = 1 \Rightarrow y = 2$

So, $P = (x_{1}, y_{1}) = (1, 2) \Rightarrow x_{1} y_{1} = 2$

Use mid-point condition

Let $Q = (x_{2}, y_{2})$ , $R = (x_{3}, y_{3})$ .

Since P is the mid-point of QR

$\Rightarrow x_{2} + x_{3} = 2 x_{1} = 2, y_{2} + y_{3} = 2 y_{1} = 4$

So, we can write : $x_{3} = 2 - x_{2}, y_{3} = 4 - y_{2}$

(b) Given

$P Q = \frac{2 \sqrt{2}}{3} \Rightarrow P Q^{2} = {(x_{2} - 1)}^{2} + {(y_{2} - 2)}^{2} = \frac{8}{9}$

Let's denote : $x_{2} = a, y_{2} = b, x_{3} = 2 - a, y_{3} = 5 - b$

${(a - 1)}^{2} + {(b - 2)}^{2} = \frac{8}{9}$

$\Rightarrow a^{2} - 2 a + 1 + b^{2} - 4 b + 4 = \frac{8}{9}$

$\Rightarrow a^{2} + b^{2} - 2 a - 4 b + 5 = \frac{8}{9}$

$\Rightarrow 9 a^{2} + 9 b^{2} - 18 a - 36 b + 37 = 0$

Hence, $a = \frac{5}{3}, b = \frac{4}{3}$

$x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} = 2 + a b + (2 - a) (4 - b)$

$9 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3}) = 9 (10 + 2 a b - 2 b - 4 a)$

= 90 + 18ab – 18b – 36a = 46.

Q 27 :

Let $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ be an ellipse. Ellipse $E_{i}' s$ are constructed such that their centres and eccentricities are same as that of $E_{1}$ , and the length of minor axis of $E_{i}$ is the length of major axis of $E_{i + 1} (i \geq 1)$ . If $A_{i}$ is the area of the ellipse $E_{i}$ , then $\frac{5}{π} (\sum_{i = 1}^{\infty} A_{i})$ , is equal to __________. [2025]

(54)

Given, $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \Rightarrow e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{\sqrt{5}}{3}$

As length of minor axis of $E_{i}$ is the length of major axis of $E_{i + 1}$ .

$E_{2} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1 \Rightarrow e = \sqrt{1 - \frac{a^{2}}{4^{2}}} = \frac{\sqrt{5}}{3}$ [ $∵$ Eccentricities are same]

$\Rightarrow a = \frac{4}{3}$ $∴ E_{2} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{4} = 1$

Now, $E_{3} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{b^{2}}{\frac{16}{9}}} = \frac{\sqrt{5}}{3}$

$\Rightarrow b^{2} = \frac{64}{81}$ $∴ E_{3} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{\frac{64}{81}} = 1$

Now, $A_{1} = π a b = π \times 3 \times 2 = 6 π$ ,

$A_{2} = π \times \frac{4}{3} \times 2 = \frac{8}{3} π$

and $A_{3} = π \times \frac{4}{3} \times \frac{8}{9} = \frac{32}{27} π$

$∴ \sum_{i = 1}^{\infty} A_{i} = 6 π + \frac{8}{3} π + \frac{32}{27} π + . . . \infty$

$= \frac{6 π}{1 - \frac{4}{9}} = \frac{54 π}{5}$ $[∵ S_{\infty} = \frac{a}{1 - r}]$

So, $\frac{5}{π} \sum_{i = 1}^{\infty} A_{i} = \frac{5}{π} \times \frac{54 π}{5} = 54$ .

Q 28 :

In a group of 100 persons, 75 speak English and 40 speak Hindi. Each person speaks at least one of the two languages. If the number of persons who speak only English is $α$ and the number of persons who speak only Hindi is $β$ , then the eccentricity of the ellipse $25 (β^{2} x^{2} + α^{2} y^{2}) = α^{2} β^{2}$ is [2023]

$\frac{\sqrt{117}}{12}$
$\frac{\sqrt{119}}{12}$
$\frac{\sqrt{129}}{12}$
$\frac{3 \sqrt{15}}{12}$

1

2

3

4

(2)

Let $γ$ be the number of persons who speak both English and Hindi.

According to the question, we have

$α + β + γ = 100 (i)$

$α + γ = 75 (ii)$

and $β + γ = 40 (iii)$

Solving (i), (ii) and (iii), we get

$α = 60,$ $β = 25$ and $γ = 15$

So, equation of the given ellipse becomes,

$25 (25^{2} x^{2} + 60^{2} y^{2}) = 60^{2} \times 25^{2}$

$\Rightarrow \frac{x^{2}}{{(60 / 5)}^{2}} + \frac{y^{2}}{{(25 / 5)}^{2}} = 1 \Rightarrow \frac{x^{2}}{12^{2}} + \frac{y^{2}}{5^{2}} = 1$

So, eccentricity of ellipse $= \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{5^{2}}{12^{2}}} = \frac{\sqrt{119}}{12}$

Q 29 :

Let the ellipse $E : x^{2} + 9 y^{2} = 9$ intersect the positive $x$ - and $y$ -axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is $\frac{m}{n}$ , where m and n are coprime, then $m - n$ is equal to [2023]

17
15
18
16

1

2

3

4

(1)

We have, $E : \frac{x^{2}}{9} + \frac{y^{2}}{1} = 1$

$A \equiv (3, 0)$

$B \equiv (0, 1)$

Equation of line passing through A(3, 0) and B(0, 1) is $\frac{y - 0}{x - 3} = \frac{1 - 0}{0 - 3} \Rightarrow x + 3 y = 3$ ...(i)

The equation of the circle with radius 3 is $x^{2} + y^{2} = 9$ ...(ii)

From (i) and (ii), we get
${(3 - 3 y)}^{2} + y^{2} = 9 \Rightarrow 10 y^{2} - 18 y = 0$ $\Rightarrow y = 0, \frac{9}{5}$

$∴$ $Area of triangle O P A = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10} = \frac{m}{n}$

$∴$ $m - n = 17$

Q 30 :

Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285 .$ Then the common tangents are inclined to the minor axis of the ellipse at the angle [2023]

$\frac{π}{12}$
$\frac{π}{4}$
$\frac{π}{6}$
$\frac{π}{3}$

1

2

3

4

(4)

$Given ellipse is \frac{x^{2}}{19} + \frac{y^{2}}{15} = 1$

$Equation of tangent to the ellipse is given by$

$y = m x \pm \sqrt{19 m^{2} + 15}$ ...(i)

$Equation of tangent to the circle x^{2} + y^{2} = 16 is given by$

$y = m x \pm 4 \sqrt{1 + m^{2}}$ ...(ii)

$From (i) and (ii), m x \pm 4 \sqrt{1 + m^{2}} = m x \pm \sqrt{19 m^{2} + 15}$

$\Rightarrow 4 \sqrt{1 + m^{2}} = \sqrt{19 m^{2} + 15}$ $\Rightarrow 16 (1 + m^{2}) = 19 m^{2} + 15$

$\Rightarrow 16 + 16 m^{2} = 19 m^{2} + 15$

$\Rightarrow 1 = 3 m^{2} \Rightarrow m^{2} = \frac{1}{3} \Rightarrow m = \pm \frac{1}{\sqrt{3}}$

$\Rightarrow \tan θ = \pm \frac{1}{\sqrt{3}} \Rightarrow θ = \frac{π}{6} with x-axis$

$∴$ $Common tangent inclined to the minor axis of the ellipse at the angle \frac{π}{3} .$

Topic Question Set

Q 2 :

Let $f (x) = x^{2} + 9$ , $g (x) = \frac{x}{x - 9}$ and a = fog (10), b = gof (3). If e and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{a} + \frac{y^{2}}{b} = 1$ , then $8 e^{2}$ + $l^{2}$ is equal to [2024]

Q 3 :

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum is $\sqrt{14}$ . Then the square of the eccentricity of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is: [2024]

Q 5 :

The length of the chord of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid point is $(1, \frac{2}{5})$ , is equal to : [2024]

Q 6 :

If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is : [2024]

Q 8 :

Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6. Let an ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola P. Then the square of the length of the latus rectum of E is [2024]

Q 9 :

If the points of intersection of two distinct conics $x^{2} + y^{2} = 4 b$ and $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ lie on the curve $y^{2} = 3 x^{2}$ , then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is _________. [2024]

Q 10 :

If S and $S'$ are the foci of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{9} = 1$ and P be a point on the ellipse, then $\min (S P \cdot S' P) + \max (S P \cdot S' P)$ is equal to : [2025]

Q 11 :

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is : [2025]

Q 12 :

A line passing through the point $P (\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$ at A and B such that (PA) $\cdot$ (PB) is maximum. Then $5 (P A^{2} + P B^{2})$ is equal to : [2025]

Q 14 :

The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, –3) and eccentricity is $\frac{4}{5}$ , is [2025]

Q 16 :

Let for two distinct values of p the lines y = x + p touch the ellipse $E : \frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$ at the points A and B. Let the line y = x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to : [2025]

Q 17 :

Let the system of equations

x + 5y – z = 1

4x + 3y – 3z = 7

24x + y + $λ$ z = $μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

Q 19 :

Let the ellipse $3 x^{2} + p y^{2} = 4$ pass through the centre C of the circle $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ of radius r. Let $f_{1}, f_{2}$ be the focal distances of the point C on the ellipse. Then $6 f_{1} f_{2} - r$ is equal to [2025]

Q 20 :

The length of the chord of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$ , whose mid-point is $(1, \frac{1}{2})$ , is: [2025]

Q 21 :

Let the product of the focal distances of the point $(\sqrt{3}, \frac{1}{2})$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , (a > b), be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipse is [2025]

Q 22 :

The equation of the chord, of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid-point is (3, 1) is : [2025]

Q 24 :

If the mid-point of a chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is $(\sqrt{2}, \frac{4}{3})$ and the length of the chord is $\frac{2 \sqrt{α}}{3}$ , then $α$ is: [2025]

Q 25 :

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ , whose mid point is $(\frac{5}{2}, \frac{1}{2})$ , then $α + β$ is equal to : [2025]

Q 30 :

Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285 .$ Then the common tangents are inclined to the minor axis of the ellipse at the angle [2023]

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Topic Question Set

Q 2 : Let f(x)=x2+9, g(x)=xx – 9 and a = fog (10), b = gof (3). If e and l denote the eccentricity and the length of the latus rectum of the ellipse x2a+y2b=1, then 8e2 + l2 is equal to [2024]

Q 3 : Let x2a2+y2b2=1, a > b be an ellipse, whose eccentricity is 12 and the length of the latusrectum is 14. Then the square of the eccentricity of x2a2–y2b2=1 is: [2024]

Q 5 : The length of the chord of the ellipse x225+y216=1, whose mid point is (1, 25), is equal to : [2024]

Q 6 : If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is : [2024]

Q 8 : Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6. Let an ellipse E : x2a2+y2b2=1, a > b of eccentricity 12 pass through the focus of the parabola P. Then the square of the length of the latus rectum of E is [2024]

Q 9 : If the points of intersection of two distinct conics x2+y2=4b and x216+y2b2=1 lie on the curve y2=3x2, then 33 times the area of the rectangle formed by the intersection points is _________. [2024]

Q 10 : If S and S' are the foci of the ellipse x218+y29=1 and P be a point on the ellipse, then min(SP·S'P)+max(SP·S'P) is equal to : [2025]

Q 11 : If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is : [2025]

Q 12 : A line passing through the point P(5,5) intersects the ellipse x236+y225=1 at A and B such that (PA)·(PB) is maximum. Then 5(PA2+PB2) is equal to : [2025]

Q 14 : The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, –3) and eccentricity is 45, is [2025]

Q 16 : Let for two distinct values of p the lines y = x + p touch the ellipse E : x242+y232=1 at the points A and B. Let the line y = x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to : [2025]

Q 17 : Let the system of equations x + 5y – z = 1 4x + 3y – 3z = 7 24x + y + λz = μ λ,μ∈R, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7≤x+y+z≤77, is : [2025]

Q 18 : Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse x216+y2n=1 is: [2025]

Q 19 : Let the ellipse 3x2+py2=4 pass through the centre C of the circle x2+y2–2x–4y–11=0 of radius r. Let f1,f2 be the focal distances of the point C on the ellipse. Then 6f1f2–r is equal to [2025]

Q 20 : The length of the chord of the ellipse x24+y22=1, whose mid-point is (1,12), is: [2025]

Q 21 : Let the product of the focal distances of the point (3,12) on the ellipse x2a2+y2b2=1, (a > b), be 74. Then the absolute difference of the eccentricities of two such ellipse is [2025]

Q 22 : The equation of the chord, of the ellipse x225+y216=1, whose mid-point is (3, 1) is : [2025]

Q 23 : Let the ellipse E1 : x2a2+y2b2=1, a > b and E2 : x2A2+y2B2=1, A < B have same eccentricity 13. Let the product of their lengths of latus rectums be 323, and the distance between the foci of E1 be 4. If E1 and E2 meet at A, B, C and D, then the area of the quadrilateral ABCD equals : [2025]

Q 24 : If the mid-point of a chord of the ellipse x29+y24=1 is (2,43) and the length of the chord is 2α3, then α is: [2025]

Q 25 : If αx+βy=109 is the equation of the chord of the ellipse x29+y24=1, whose mid point is (52,12), then α+β is equal to : [2025]

Q 30 : Let a circle of radius 4 be concentric to the ellipse 15x2+19y2=285. Then the common tangents are inclined to the minor axis of the ellipse at the angle [2023]

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Q 2 :

Let $f (x) = x^{2} + 9$ , $g (x) = \frac{x}{x - 9}$ and a = fog (10), b = gof (3). If e and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{a} + \frac{y^{2}}{b} = 1$ , then $8 e^{2}$ + $l^{2}$ is equal to [2024]

Q 3 :

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum is $\sqrt{14}$ . Then the square of the eccentricity of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is: [2024]

Q 5 :

The length of the chord of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid point is $(1, \frac{2}{5})$ , is equal to : [2024]

Q 6 :

If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is : [2024]

Q 8 :

Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6. Let an ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola P. Then the square of the length of the latus rectum of E is [2024]

Q 9 :

If the points of intersection of two distinct conics $x^{2} + y^{2} = 4 b$ and $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ lie on the curve $y^{2} = 3 x^{2}$ , then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is _________. [2024]

Q 10 :

If S and $S'$ are the foci of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{9} = 1$ and P be a point on the ellipse, then $\min (S P \cdot S' P) + \max (S P \cdot S' P)$ is equal to : [2025]

Q 11 :

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is : [2025]

Q 12 :

A line passing through the point $P (\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$ at A and B such that (PA) $\cdot$ (PB) is maximum. Then $5 (P A^{2} + P B^{2})$ is equal to : [2025]

Q 14 :

The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, –3) and eccentricity is $\frac{4}{5}$ , is [2025]

Q 16 :

Let for two distinct values of p the lines y = x + p touch the ellipse $E : \frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$ at the points A and B. Let the line y = x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to : [2025]

Q 17 :

Let the system of equations

x + 5y – z = 1

4x + 3y – 3z = 7

24x + y + $λ$ z = $μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

Q 19 :

Let the ellipse $3 x^{2} + p y^{2} = 4$ pass through the centre C of the circle $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ of radius r. Let $f_{1}, f_{2}$ be the focal distances of the point C on the ellipse. Then $6 f_{1} f_{2} - r$ is equal to [2025]

Q 20 :

The length of the chord of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$ , whose mid-point is $(1, \frac{1}{2})$ , is: [2025]

Q 21 :

Let the product of the focal distances of the point $(\sqrt{3}, \frac{1}{2})$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , (a > b), be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipse is [2025]

Q 22 :

The equation of the chord, of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid-point is (3, 1) is : [2025]

Q 24 :

If the mid-point of a chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is $(\sqrt{2}, \frac{4}{3})$ and the length of the chord is $\frac{2 \sqrt{α}}{3}$ , then $α$ is: [2025]

Q 25 :

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ , whose mid point is $(\frac{5}{2}, \frac{1}{2})$ , then $α + β$ is equal to : [2025]

Q 30 :

Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285 .$ Then the common tangents are inclined to the minor axis of the ellipse at the angle [2023]