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

Q 21 :
Let the shortest distance from (a, 0), a > 0, to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is: [2025]

$x^{2} + y^{2} - 4 x + 3 = 0$
$x^{2} + y^{2} - 8 x + 7 = 0$
$x^{2} + y^{2} - 10 x + 9 = 0$
$x^{2} + y^{2} - 6 x + 5 = 0$

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(4)

Equation of normal at $P (t^{2}, 2 t)$ is given by

$y + t x = 2 t + t^{3}$ ... (i)

Put x = a, y = 0 in equation (i), we get

$a t = 2 t + t^{3}$

$\Rightarrow a = 2 + t^{2}$

$∴$ The point Q is $(2 + t^{2}, 0)$

$\Rightarrow P Q = 4 \Rightarrow 4 + 4 t^{2} = 16$

$\Rightarrow 4 t^{2} = 12 \Rightarrow t^{2} = 3$

$∴ a = 5 and Q \equiv (5, 0)$

$∵$ Focus of parabola is (1, 0) and centre of circle lie on axis of parabola.
$\Rightarrow$ (1, 0) and (5, 0) will be the end points of diameter of the circle.

$∴$ Equation of circle is $(x - 1) (x - 5) + y^{2} = 0$

$\Rightarrow x^{2} + y^{2} - 6 x + 5 = 0$ .

Q 22 :
If the equation of the parabola with vertex $V (\frac{3}{2}, 3)$ and the directrix x + 2y = 0 is $α x^{2} + β y^{2} - γ x y - 30 x - 60 y + 225 = 0$ , then $α + β + γ$ is equal to : [2025]

9
6
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7

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(1)

Given : Vertex of parabola $(\frac{3}{2}, 3)$ and directrix is x + 2y = 0

Since, axis is $⊥$ to directrix and passes through vertex, then equation of axis

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

$∴$ Foot of directrix is intersecting point of

y = 2x & 2y + x = 0 i.e., (0, 0)

$∴$ Focus $\equiv$ (3, 6)

Using definition of parabola,

$P S^{2} = P M^{2}$

$\Rightarrow {(x - 3)}^{2} + {(y - 6)}^{2} = {(\frac{x + 2 y}{\sqrt{5}})}^{2}$

$\Rightarrow x^{2} + 9 - 6 x + y^{2} + 36 - 12 y = \frac{x^{2} + 4 y^{2} + 4 x y}{5}$

$\Rightarrow 4 x^{2} + y^{2} - 4 x y - 30 x - 60 y + 225 = 0$

On comparing we get $α = 4, β = 1 and γ = 4$

Hence, $α + β + γ$ = 4 + 1 + 4 = 9.

Q 23 :
Let ABCD be a trapezium whose vertices lie on the parabola $y^{2} = 4 x$ . Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length $\frac{25}{4}$ and it passes through the point (1, 0), then the area of ABCD is [2025]

$\frac{25}{2}$
$\frac{125}{8}$
$\frac{75}{4}$
$\frac{75}{8}$

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(3)

Let $A (a t_{1}^{2}, 2 a t_{1})$ and $C (\frac{a}{t_{1}^{2}}, \frac{- 2 a}{t_{1}})$ be the points lies on parabola $y^{2} = 4 x$ .

$∴$ Length of $A C = a {(t_{1} + \frac{1}{t_{1}})}^{2} = \frac{25}{4}$ (Given)

$\Rightarrow t_{1} + \frac{1}{t_{1}} = \pm \frac{5}{2}$

So, $A (\frac{1}{4}, 1), B (4, 4), C (4, - 4)$ and $D (\frac{1}{4}, - 1)$

$∴$ The area of trapezium ABCD = $\frac{1}{2} (8 + 2) (4 - \frac{1}{4}) = \frac{75}{4}$ .

Q 24 :
Two parabolas have the same focus (4, 3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then ${(A B)}^{2}$ is equal to : [2025]

392
96
384
192

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(4)

Let the two parabolas intersect at $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ .

$∴$ Equation of parabolas are

${(x - 4)}^{2} + {(y - 3)}^{2} = x^{2}$ ... (i)

and ${(x - 4)}^{2} + {(y - 3)}^{2} = y^{2}$ ... (ii)

From (i) and (ii), we get x = y

$\Rightarrow x^{2} - 14 x + 25 = 0$ and

$\Rightarrow x_{1} + x_{2} = 14 and x_{1} x_{2} = 25$

$∴ A B^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} = 2 {(x_{1} - x_{2})}^{2}$

$= 2 {[(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2}] = 2 [14^{2} - 100] = 192$ .

Q 25 :
The focus of the parabola $y^{2} = 4 x + 16$ is the centre of the circle C of radius 5. If the values of $λ$ , for which C passes through the point of intersection of the lines 3x – y = 0 and x + $λ$ y = 4, are $λ_{1}$ and $λ_{2}, λ_{1} < λ_{2}$ , then $12 λ_{1} + 29 λ_{2}$ is equal to __________. [2025]

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(15)

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

Focus of parabola = (–3, 0) $\Rightarrow$ Center $\equiv$ (–3, 0)

$∴$ Equation of circle is given by ${(x + 3)}^{2} + y^{2} = 25$

Intersection point of 3x – y = 0 and x + $λ$ y = 4 is

$(\frac{4}{3 λ + 1}, \frac{12}{3 λ + 1})$

Circle passes through the point of intersection of two lines 3x – y = 0 and x + $λ$ y = 4.

$\Rightarrow 144 λ^{2} + 24 λ - 168 = 0 \Rightarrow 18 λ^{2} + 3 λ - 21 = 0$

$\Rightarrow λ = - \frac{7}{6}, 1$

Now, $12 λ_{1} + 29 λ_{2}$ = –14 + 29 = 15.

Q 26 :
Let A and B be the two points of intersection of the line y + 5 = 0 and the mirror image of the parabola $y^{2} = 4 x$ with respect to the line x + y + 4 = 0. If d denotes the distance between A and B, and $a$ denotes the area of $△$ SAB, where S is the focus of the parabola $y^{2} = 4 x$ , then the value of (a + d) is __________. [2025]

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(14)

Image of point (0, 0) w.r.t. to line x + y + 4 = 0

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

$\frac{x - 0}{1} = \frac{y - 0}{1} = \frac{- 2 (4)}{2} \Rightarrow x = y = - 4$

Image of focus (1, 0) w.r.t. to line x + y + 4 = 0

$\frac{x - 1}{1} = \frac{y - 0}{1} = - 2 \frac{(1 + 4)}{2} x = - 4; y = - 5$

Equation of mirror image of parabola

${(x - h)}^{2} = - 4 a (y - k) {(x + 4)}^{2} = - 4 (1) (y + 4)$

Put y = –5; we get x = –6 and –2

$∴$ A = (–6, –5); B = (–2, –5)

Distance between the points, d = AB = 4

Area of $△$ SAB, $a = \frac{1}{2} \times 4 \times 5 = 10$

So, a + d = 14.

Q 27 :
Let $y^{2} = 12 x$ be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ) = $\frac{147}{4}$ . Let C be the circle described taking PQ as a diameter. If the equation of a circle C is $64 x^{2} + 64 y^{2} - α x - 64 \sqrt{3} y = β$ , then $β - α$ is equal to __________. [2025]

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(1328)

Given, equation of parabola is $y^{2} = 12 x$

Focus = S = (3, 0)

Let $P (3 t_{1}^{2}, 6 t_{1})$ and $Q (3 t_{2}^{2}, 6 t_{2})$ are points on parabola

Also, $t_{1} t_{2} = - 1$

$∴ P \equiv (3 t^{2}, 6 t) and Q \equiv (\frac{3}{t^{2}}, \frac{- 6}{t})$

Now, $(S P) (S Q) = \frac{147}{4} \Rightarrow (3 + 3 t^{2}) (3 + \frac{3}{t^{2}}) = \frac{147}{4}$

( $∵$ (SP) = PM and (SQ) = QN, where PM and QN are perpendicular distance from directrix)

$\Rightarrow \frac{{(1 + t^{2})}^{2}}{t^{2}} = \frac{49}{12} \Rightarrow 12 t^{4} - 25 t^{2} + 12 = 0$

$\Rightarrow t^{2} = \frac{3}{4}, \frac{4}{3} \Rightarrow t = \pm \frac{\sqrt{3}}{2} or t = \pm \frac{2}{\sqrt{3}}$

Case I : When $t = \frac{\sqrt{3}}{2} or t = - \frac{2}{\sqrt{3}}$

Points are $P (\frac{9}{4}, 3 \sqrt{3}) and Q (4, - 4 \sqrt{3})$

$∴$ Equation of circle is

$(x - 4) (x - \frac{9}{4}) + (y - 3 \sqrt{3}) (y + 4 \sqrt{3}) = 0$

$\Rightarrow x^{2} + y^{2} - \frac{25 x}{4} + \sqrt{3} y - 27 = 0$

On comparing with given equation of circle $64 x^{2} + 64 y^{2} - α x$

$- 64 \sqrt{3} y = β$ , we get $α$ = 400, $β$ = 1728

Case II : When $t = \frac{- \sqrt{3}}{2} or t = \frac{2}{\sqrt{3}}$

Point are $P (\frac{9}{4}, - 3 \sqrt{3}) and Q (4, 4 \sqrt{3})$

Similarly, we get $α$ = 400, $β$ = 1728

$∴ β - α$ = 1728 – 400 = 1328.

Topic Question Set

Q 21 :
Let the shortest distance from (a, 0), a > 0, to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is: [2025]

Q 22 :
If the equation of the parabola with vertex $V (\frac{3}{2}, 3)$ and the directrix x + 2y = 0 is $α x^{2} + β y^{2} - γ x y - 30 x - 60 y + 225 = 0$ , then $α + β + γ$ is equal to : [2025]

Q 23 :
Let ABCD be a trapezium whose vertices lie on the parabola $y^{2} = 4 x$ . Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length $\frac{25}{4}$ and it passes through the point (1, 0), then the area of ABCD is [2025]

Q 24 :
Two parabolas have the same focus (4, 3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then ${(A B)}^{2}$ is equal to : [2025]

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Topic Question Set

Q 21 : Let the shortest distance from (a, 0), a > 0, to the parabola y2=4x be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is: [2025]

Q 22 : If the equation of the parabola with vertex V(32,3) and the directrix x + 2y = 0 is αx2+βy2–γxy–30x–60y+225=0, then α+β+γ is equal to : [2025]

Q 23 : Let ABCD be a trapezium whose vertices lie on the parabola y2=4x. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length 254 and it passes through the point (1, 0), then the area of ABCD is [2025]

Q 24 : Two parabolas have the same focus (4, 3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then (AB)2 is equal to : [2025]

Q 25 : The focus of the parabola y2=4x+16 is the centre of the circle C of radius 5. If the values of λ, for which C passes through the point of intersection of the lines 3x – y = 0 and x + λy = 4, are λ1 and λ2,λ1<λ2, then 12λ1+29λ2 is equal to __________. [2025]

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Q 27 : Let y2=12x be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ) = 1474. Let C be the circle described taking PQ as a diameter. If the equation of a circle C is 64x2+64y2–αx–643y=β, then β–α is equal to __________. [2025]

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Q 21 :
Let the shortest distance from (a, 0), a > 0, to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is: [2025]

Q 22 :
If the equation of the parabola with vertex $V (\frac{3}{2}, 3)$ and the directrix x + 2y = 0 is $α x^{2} + β y^{2} - γ x y - 30 x - 60 y + 225 = 0$ , then $α + β + γ$ is equal to : [2025]

Q 23 :
Let ABCD be a trapezium whose vertices lie on the parabola $y^{2} = 4 x$ . Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length $\frac{25}{4}$ and it passes through the point (1, 0), then the area of ABCD is [2025]

Q 24 :
Two parabolas have the same focus (4, 3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then ${(A B)}^{2}$ is equal to : [2025]