JEE Advanced Maths PYQ – Conic Sections (Parabola) | Previous Year Questions with Solutions

Q 1 :

Let $P$ be a point on the parabola $y^{2} = 4 a x$ , where $a > 0$ . The normal to the parabola at $P$ meets the $x$ -axis at a point $Q$ . The area of the triangle $P F Q$ , where $F$ is the focus of the parabola, is 120. If the slope $m$ of the normal and $a$ are both positive integers, then the pair $(a, m)$ is [2023]

(2, 3)
(1, 3)
(2, 4)
(3, 4)

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2

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4

(1)

$Let point P (a m^{2}, - 2 a m)$

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$Equation of normal at P (a m^{2}, - 2 a m) is$

$y = m x - 2 a m - a m^{3}$

$Area of ∆ P F Q$

$= \frac{1}{2} (a + a m^{2}) \times 2 a m = 120$

$\Rightarrow a^{2} m (1 + m^{2}) = 120$

$Pair (a, m) = (2, 3) satisfies the above equation.$

Q 2 :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from (0, 0) to $(x, y)$ in the ratio 1 : 3. Then the locus of $P$ is [2011]

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

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4

(3)

$Let A (x, y) = (t^{2}, 2 t) be any point on parabola y^{2} = 4 x .$

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$Let P (h, k) divide O A in the ratio 1 : 3$

$∴$ $(h, k) = (\frac{t^{2}}{4}, \frac{2 t}{4}) \Rightarrow h = \frac{t^{2}}{4}, k = \frac{t}{2} \Rightarrow h = k^{2}$

$∴$ $Locus of P (h, k) is x = y^{2}$

Q 3 :

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $\sqrt{2}$ and $2 \sqrt{2}$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

${(x + y)}^{2} = (x - y - 2)$
${(x - y)}^{2} = (x + y - 2)$
${(x - y)}^{2} = 4 (x + y - 2)$
${(x - y)}^{2} = 8 (x + y - 2)$

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2

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4

(4)

Since distance of vertex and focus of the parabola from origin is $\sqrt{2}$ and $2 \sqrt{2}$ ,

$∴$ Vertex is (1, 1) and focus is (2, 2), directrix $x + y = 0$

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$∴$ Equation of parabola is

${(x - 2)}^{2} + {(y - 2)}^{2} = {(\frac{x + y}{\sqrt{2}})}^{2}$

$\Rightarrow 2 (x^{2} - 4 x + 4) + 2 (y^{2} - 4 y + 4) = x^{2} + y^{2} + 2 x y$

$\Rightarrow x^{2} + y^{2} - 2 x y = 8 (x + y - 2) \Rightarrow {(x - y)}^{2} = 8 (x + y - 2)$

Q 4 :

Tangent to the curve $y = x^{2} + 6$ at a point (1, 7) touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are [2005]

$(- 6, - 11)$
$(- 9, - 13)$
$(- 10, - 15)$
$(- 6, - 7)$

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4

(4)

The given curve is $y = x^{2} + 6$

Equation of tangent at (1, 7) is $\frac{1}{2} (y + 7) = x \cdot 1 + 6 \Rightarrow 2 x - y + 5 = 0 ...(i)$

Since tangent (i) touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ with centre $C (- 8, - 6)$ at Q.

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$∴$ Equation of $C Q$ which is perpendicular to (i) is

$y + 6 = - \frac{1}{2} (x + 8) \Rightarrow x + 2 y + 20 = 0 ...(ii)$

On solving equations (i) and (ii), we get the coordinates of $Q$ as $x = - 6, y = - 7$

$∴$ Coordinates of $Q$ is $(- 6, - 7)$

Q 5 :

The angle between the tangents drawn from the point (1, 4) to the parabola $y^{2} = 4 x$ is [2004]

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{2}$

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4

(3)

If $m$ be the slope of the tangent to the parabola, then its equation is $y = m x + \frac{1}{m}$

Since the tangent passes through (1, 4)

$∴$ $4 = m + \frac{1}{m} \Rightarrow m^{2} - 4 m + 1 = 0$

If the angle between two tangents to the parabola be $θ$ , then

$\tan θ =$ $| \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$ $= \frac{\sqrt{{(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2}}}{1 + m_{1} m_{2}}$

$= \frac{\sqrt{16 - 4}}{1 + 1} = \sqrt{3} \Rightarrow θ = \frac{π}{3}$

Topic Question Set

Q 2 :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from (0, 0) to $(x, y)$ in the ratio 1 : 3. Then the locus of $P$ is [2011]

Q 3 :

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $\sqrt{2}$ and $2 \sqrt{2}$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

Q 4 :

Tangent to the curve $y = x^{2} + 6$ at a point (1, 7) touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are [2005]

Q 5 :

The angle between the tangents drawn from the point (1, 4) to the parabola $y^{2} = 4 x$ is [2004]

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Topic Question Set

Q 1 : Let P be a point on the parabola y2=4ax, where a>0. The normal to the parabola at P meets the x-axis at a point Q. The area of the triangle PFQ, where F is the focus of the parabola, is 120. If the slope m of the normal and a are both positive integers, then the pair (a,m) is [2023]

Q 2 : Let (x,y) be any point on the parabola y2=4x. Let P be the point that divides the line segment from (0, 0) to (x,y) in the ratio 1 : 3. Then the locus of P is [2011]

Q 3 : The axis of a parabola is along the line y=x and the distances of its vertex and focus from origin are 2 and 22 respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

Q 4 : Tangent to the curve y=x2+6 at a point (1, 7) touches the circle x2+y2+16x+12y+c=0 at a point Q. Then the coordinates of Q are [2005]

Q 5 : The angle between the tangents drawn from the point (1, 4) to the parabola y2=4x is [2004]

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Q 2 :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from (0, 0) to $(x, y)$ in the ratio 1 : 3. Then the locus of $P$ is [2011]

Q 3 :

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $\sqrt{2}$ and $2 \sqrt{2}$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

Q 4 :

Tangent to the curve $y = x^{2} + 6$ at a point (1, 7) touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are [2005]

Q 5 :

The angle between the tangents drawn from the point (1, 4) to the parabola $y^{2} = 4 x$ is [2004]