Parabola JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Parabola, free to read — no sign-in needed. The full chapter has 35 questions; sign in to attempt the remaining 30 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $T$ be the tangent to the parabola $y^2 = 16x$ at the point $(64, 32)$. Let $L$ be the tangent to the same parabola at another point $(x_1, y_1)$ on the parabola. If $L$ and $T$ are perpendicular to each other, then the distance between the point $(x_1, y_1)$ and the focus of the parabola, is
   1. A.$\dfrac{15}{4}$
   2. B.$4$
   3. C.$\dfrac{17}{4}$
   4. D.$5$
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   Answer: (C)

   The equation of the parabola is $y^2 = 16x$. Comparing with $y^2 = 4ax$, we get $a = 4$. Let the parametric coordinates of the point $(64, 32)$ be $(at_1^2, 2at_1)$. $2at_1 = 32 \Rightarrow 8t_1 = 32 \Rightarrow t_1 = 4$ The slope of the tangent $T$ at $t_1$ is $m_1 = \dfrac{1}{t_1} = \dfrac{1}{4}$. Let the tangent $L$ be at the point $(x_1, y_1)$ with parameter $t_2$. Its slope is $m_2 = \dfrac{1}{t_2}$. Since $T$ and $L$ are perpendicular, $m_1 m_2 = -1$. $\dfrac{1}{4} \times \dfrac{1}{t_2} = -1 \Rightarrow t_2 = -\dfrac{1}{4}$ The $x$-coordinate of the point $(x_1, y_1)$ is $x_1 = at_2^2 = 4\left(-\dfrac{1}{4}\right)^2 = \dfrac{1}{4}$. The distance of a point $(x_1, y_1)$ on the parabola from the focus is given by its focal distance $x_1 + a$. Distance $= \dfrac{1}{4} + 4 = \dfrac{17}{4}$ Answer: $\dfrac{17}{4}$
2. Q2JEE Advanced Adv 2017 (Paper 1)
   If a chord, which is not a tangent, of the parabola ${y}^{2}=16x$ has the equation $2x+y=p$, and midpoint (h, k), then which of the following is (are) possible value(s) of $p,h$ and $k$?
   1. A.$p=-2,h=2,k=-4$
   2. B.$p=5,h=4,k=-3$
   3. C.$p=-1,h=1,k=-3$
   4. D.$p=2,h=3,k=-4$
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   Answer: (D)

   Equation of chord with mid point $(h,k)$: $k.y-16\left(\dfrac{x+h}{2}\right)={k}^{2}-16h$ $\Rightarrow 8x-ky+{k}^{2}-8h=0$ Comparing with $2x+y-p=0,$ we get $k=-4;2h-p=4$ Only $p=2,h=3,k=-4$ satisfies above relation.
3. Q3JEE Advanced Adv 2015 (Paper 1)
   If the normal of the parabola ${y}^{2}=4x$ drawn at the end points of its latus rectum are tangents to the circle ${\left(x-3\right)}^{2}+{\left(y+2\right)}^{2}={r}^{2},$ then the value of ${r}^{2}$ is
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   Answer: 2

   End points of latus rectum $A(1,2),B(1,–2)$ Equation of normal at $A\left[\dfrac{dy}{dx}=1\right]$ $y-2=-1\left(x-1\right)$ $x+y-3=0$ ...(i) $r=$ Distance of line (i) from $\left(3,-2\right)$ $r=\left|\dfrac{3-2-3}{\sqrt{2}}\right|=\sqrt{2}$ ${r}^{2}=2$
4. Q4JEE Advanced Adv 2014 (Paper 2)
   Paragraph: Let $a, r, s, t$ be nonzero real numbers. Let $P\left(a t^{2}, 2 a t\right), Q, R\left(a r^{2}, 2 a r\right)$ and $S\left(a s^{2}, 2 a s\right)$ be distinct points on the parabola $y^{2}=4 a x$. Suppose that $P Q$ is the focal chord and lines $Q R$ and $P K$ are parallel, where $K$ is the point $(2 a, 0)$. Question: If $s t=1$, then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is
   1. A.$\dfrac{{\left({t}^{2}+1\right)}^{2}}{2{t}^{3}}$
   2. B.$\dfrac{a{\left({t}^{2}+1\right)}^{2}}{2{t}^{3}}$
   3. C.$\dfrac{a{\left({t}^{2}+1\right)}^{2}}{{t}^{3}}$
   4. D.$\dfrac{a{\left({t}^{2}+2\right)}^{2}}{{t}^{3}}$
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   Answer: (B)

   Tangent at P: ty $=x+a{t}^{2}ory=\dfrac{x}{t}+at$ Normal at S: $y+\dfrac{x}{t}=\dfrac{2a}{t}+\dfrac{a}{{t}^{3}}$ Solving, $2y=at+\dfrac{2a}{t}+\dfrac{a}{{t}^{3}}$ $y=\dfrac{a{\left({t}^{2}+1\right)}^{2}}{2{t}^{3}}$
5. Q5JEE Advanced Adv 2014 (Paper 2)
   Paragraph: Let $a, r, s, t$ be nonzero real numbers. Let $P\left(a t^{2}, 2 a t\right), Q, R\left(a r^{2}, 2 a r\right)$ and $S\left(a s^{2}, 2 a s\right)$ be distinct points on the parabola $y^{2}=4 a x$. Suppose that $P Q$ is the focal chord and lines $Q R$ and $P K$ are parallel, where $K$ is the point $(2 a, 0)$. Question: The value of $r$ is
   1. A.$-\dfrac{1}{t}$
   2. B.$\dfrac{{t}^{2}+1}{t}$
   3. C.$\dfrac{1}{t}$
   4. D.$\dfrac{{t}^{2}-1}{t}$
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   Answer: (D)

   Slope (QR) = slope (PK) $\dfrac{2at-0}{a{t}^{2}-2a}=\dfrac{-\dfrac{2a}{t}-2ar}{\dfrac{a}{{t}^{2}}-a{r}^{2}}$ $\Rightarrow \dfrac{t}{{t}^{2}-2}=-\left(\dfrac{\dfrac{1}{t}+r}{\dfrac{1}{{t}^{2}}-{r}^{2}}\right)\Rightarrow r=\dfrac{{t}^{2}-1}{t}$