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

(4)

Equation of circle with AB as diameter is

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

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

[Figure]

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

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

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

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

1

2

3

4

(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 of the circle

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

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)

[Figure]

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]

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

1

2

3

4

(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.

[Figure]

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

$\Rightarrow \frac{72}{25} + \frac{256}{25} = b^{2} \Rightarrow b^{2} = \frac{328}{25} ∴ a^{2} = \frac{2 \times 328}{25} = \frac{565}{25}$

$∴$ Length of latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 328}{25 \times \frac{\sqrt{656}}{5}} = \frac{656 \times 5}{25 \times \sqrt{656}} = \frac{\sqrt{565}}{5}$

$∴$ 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]

0%

(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 $= 3 \sqrt{3} \times 48 \sqrt{3} = 432$ .

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]

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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]

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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]