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 72 questions; sign in to attempt the remaining 67 in the exam simulator.

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.$\dfrac{12}{13}$
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   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>
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   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 _______.
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   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
   For $a\gt 0,$ let the curves ${C}_{1}:{y}^{2}=ax$ and ${C}_{2}:{x}^{2}=ay$ intersect at origin $O$ and a point $P.$ Let the line $x=b\left(0\lt b\lt a\right)$ intersect the chord $OP$ and the $x$ -axis at points $Q$ and $R,$ respectively. If the line $x=b$ bisects the area bounded by the curves, ${C}_{1}$ and ${C}_{2},$ and the area of $∆OQR=\dfrac{1}{2},$ then ‘ $a$ ’ satisfies the equation:
   1. A.${x}^{6}-6{x}^{3}+4=0$
   2. B.${x}^{6}-12{x}^{3}+4=0$
   3. C.${x}^{6}+6{x}^{3}-4=0$
   4. D.${x}^{6}-12{x}^{3}-4=0$
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   Answer: (B)

   $\int _{0}^{b}\left(\sqrt{ax}-\dfrac{{x}^{2}}{a}\right)dx=\dfrac{{a}^{2}}{6}$ $\Rightarrow \dfrac{2}{3}\sqrt{a}{b}^{\dfrac{3}{2}}-\dfrac{{b}^{3}}{3a}=\dfrac{{a}^{2}}{6}...\left(1\right)$ Also area of $\Delta OQR=\dfrac{1}{2}$ $\dfrac{1}{2}{b}^{2}=\dfrac{1}{2}\Rightarrow b=1$ Put in $\left(1\right)$ $\Rightarrow 4a\sqrt{a}-2={a}^{3}$ $\Rightarrow {a}^{6}+4{a}^{3}+4=16{a}^{3}$ $\Rightarrow {a}^{6}-12{a}^{3}+4=0$
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.$x y=\frac{64}{105}$
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   Answer: (A)

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