Parabola JEE Main previous year questions with solutions

5 solved JEE Main questions on Parabola, free to read — no sign-in needed. The full chapter has 68 questions; sign in to attempt the remaining 63 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (06 Apr, Shift 1)Chord of parabola
   Let chord PQ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points $P$ and $Q$ are in the ratio $1:2$. If the chord PQ subtends an angle $\alpha$ at the focus of the parabola, then $\sin\alpha$ is equal to:
   1. A.$\dfrac{3}{5}$
   2. B.$\dfrac{4}{5}$
   3. C.$\dfrac{5}{13}$
   4. D.$\dfrac{12}{13}$
   Show answer & solution

   Answer: (A)

   Let the coordinates of points $P$ and $Q$ on the parabola $y^2 = 12x$ be $(x_1, y_1)$ and $(x_2, y_2)$. Given that the ordinates are in the ratio $1:2$, we have $y_2 = 2y_1$. Since $P$ and $Q$ lie on the parabola, their abscissae are $x_1 = \dfrac{y_1^2}{12}$ and $x_2 = \dfrac{y_2^2}{12} = \dfrac{4y_1^2}{12} = \dfrac{y_1^2}{3}$. The length of the chord $PQ$ is $3\sqrt{13}$, so $PQ^2 = 117$. Using the distance formula: $(x_2 - x_1)^2 + (y_2 - y_1)^2 = 117$ $\left(\dfrac{y_1^2}{3} - \dfrac{y_1^2}{12}\right)^2 + (2y_1 - y_1)^2 = 117$ $\left(\dfrac{y_1^2}{4}\right)^2 + y_1^2 = 117$ $\dfrac{y_1^4}{16} + y_1^2 - 117 = 0$ $y_1^4 + 16y_1^2 - 1872 = 0$ $(y_1^2 + 52)(y_1^2 - 36) = 0$ Since $y_1^2 \gt 0$, we get $y_1^2 = 36$, which gives $y_1 = 6$ (taking the positive root by symmetry). Substituting $y_1 = 6$, we get $x_1 = \dfrac{36}{12} = 3$. Thus, $P$ is $(3, 6)$. For $Q$, $y_2 = 12$ and $x_2 = \dfrac{144}{12} = 12$. Thus, $Q$ is $(12, 12)$. The focus of the parabola $y^2 = 12x$ is $S(3, 0)$. The vectors from the focus $S$ to points $P$ and $Q$ are: $\overrightarrow{SP} = (3 - 3)\hat{i} + (6 - 0)\hat{j} = 6\hat{j}$ $\overrightarrow{SQ} = (12 - 3)\hat{i} + (12 - 0)\hat{j} = 9\hat{i} + 12\hat{j}$ The angle $\alpha$ subtended by $PQ$ at the focus is the angle between $\overrightarrow{SP}$ and $\overrightarrow{SQ}$. $\cos\alpha = \dfrac{\overrightarrow{SP} \cdot \overrightarrow{SQ}}{|\overrightarrow{SP}| |\overrightarrow{SQ}|}$ $\cos\alpha = \dfrac{0(9) + 6(12)}{6 \times \sqrt{9^2 + 12^2}} = \dfrac{72}{6 \times 15} = \dfrac{72}{90} = \dfrac{4}{5}$ Therefore, $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \dfrac{3}{5}$ Answer: $\dfrac{3}{5}$
2. Q2JEE Main 2025 (29 Jan, Shift 1)standard equation of parabola
   <p>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 :</p>
   1. A.392
   2. B.<p>384</p>
   3. C.192
   4. D.<p>96</p>
   Show answer & solution

   Answer: (C)

   <p>The parabolas are $(x-4)^2+(y-3)^2=x^2$ ...(i) and $(x-4)^2+(y-3)^2=y^2$... If point of intersection are $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ By solving (i) and (ii), we get $\begin{aligned} &amp; x_1+x_2=14 \text { and } x_1 x_2=25 \\ &amp; (A B)^2=2\left(\left(x_1+x_2\right)^2-4 x_1 x_2\right)=192 \end{aligned}$</p>
3. Q3JEE Main 2024 (29 Jan, Shift 2)Chord with given Middle Point
   Let $P(\alpha ,\beta )$ be a point on the parabola ${y}^{2}=4x$. If $P$ also lies on the chord of the parabola ${x}^{2}=8y$ whose mid point is $\left(1,\dfrac{5}{4}\right)$, then $(\alpha -28)(\beta -8)$ is equal to _______.
   Show answer & solution

   Answer: 192

   Given, Parabola ${x}^{2}=8y$ We know that, Chord with mid point $\left({x}_{1},{y}_{1}\right)$ is $T={S}_{1}$ $x{x}_{1}-8\left(\dfrac{y+{y}_{1}}{2}\right)={x}_{1}-28{y}_{1}$ $\Rightarrow x{x}_{1}-4\left(y+{y}_{1}\right)={x}_{1}-28{y}_{1}$ Now, putting the point $\left({x}_{1},{y}_{1}\right)=\left(1,\dfrac{5}{4}\right)$ we get, $\Rightarrow x-4\left(y+\dfrac{5}{4}\right)=1-8\times \dfrac{5}{4}=-9$ $∴x-4y+4=0\ldots \left(i\right)$ $(\alpha ,\beta )$ lies on (i) and also on ${y}^{2}=4x$ $∴\alpha -4\beta +4=0\ldots \left(ii\right)$ And ${\beta }^{2}=4\alpha \ldots \left(iii\right)$ Solving (ii) and (iii) ${\beta }^{2}=4(4\beta -4)$ $\Rightarrow {\beta }^{2}-16\beta +16=0$ $\Rightarrow \beta =8\pm 4\sqrt{3}$ And $\alpha =4\beta -4=28\pm 16\sqrt{3}$ $∴\left(\alpha ,\beta \right)=\left(28+16\sqrt{3},8+4\sqrt{3}\right)$ and $(28-16\sqrt{3},8-4\sqrt{3})$ Now, solving $(\alpha -28)(\beta -8)=(16\sqrt{3})(4\sqrt{3})$ $\Rightarrow (\alpha -28)(\beta -8)=192$
4. Q4JEE Main 2020 (08 Jan, Shift 1)Locus
   The locus of a point which divides the line segment joining the point $\left(0,-1\right)$ and a point on the parabola ${x}^{2}=4y$ internally in the ratio $1:2$ is:
   1. A.$9{x}^{2}-12y=8$
   2. B.$9{x}^{2}-3y=2$
   3. C.${x}^{2}-3y=2$
   4. D.$4{x}^{2}-3y=2$
   Show answer & solution

   Answer: (A)

   Let point $P$ be $\left(2t,{t}^{2}\right)$ and $Q$ be $\left(h,k\right)$ . $h=\dfrac{2t}{3},k=\dfrac{-2+{t}^{2}}{3}$ Hence, locus is $3k+2={\left(\dfrac{3h}{2}\right)s}^{2}\Rightarrow 9{x}^{2}=12y+8$
5. Q5JEE Main 2006General Equation of 2nd degree curve
   The locus of the vertices of the family of parabolas $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$ is
   1. A.$x y=\frac{105}{64}$
   2. B.$x y=\frac{3}{4}$
   3. C.$x y=\frac{35}{16}$
   4. D.$x y=\frac{64}{105}$
   Show answer & solution

   Answer: (A)

   Parabola: $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$ Vertex: $(\alpha, \beta)$ $$\begin{aligned} & \alpha=\frac{-a^2 / 2}{2 a^3 / 3}=-\frac{3}{4 a}, \beta=\frac{-\left(\frac{a^4}{4}+4 \cdot \frac{a^3}{3} \cdot 2 a\right)}{4 \frac{a^3}{3}}=-\frac{-\left(\frac{1}{4}+\frac{8}{3}\right) a^4}{\frac{4}{3} a^3} \\ & =-\frac{35}{12} \frac{a}{4} \times 3=-\frac{35}{16} a \\ & \alpha \beta=-\frac{3}{4 a}\left(-\frac{35}{16}\right) a=\frac{105}{64} . \end{aligned}$$