JEE Main 2026 — Parabola Question with Solution

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 $x_1=\frac{y_1^2}{12}$ and $x_2=\frac{y_2^2}{12}=\frac{4y_1^2}{12}=\frac{y_1^2}{3}$.

The length of the chord $PQ$ is $3\sqrt{13}$, so $P{Q}^2=117$.

Using the distance formula:

$(x_2-x_1{)}^2+(y_2-y_1{)}^2=117$

${\left(\frac{y_1^2}{3}-\frac{y_1^2}{12}\right)}^2+(2y_1-y_1{)}^2=117$

${\left(\frac{y_1^2}{4}\right)}^2+y_1^2=117$

$\frac{y_1^4}{16}+y_1^2-117=0$

$y_1^4+16y_1^2-1872=0$

$(y_1^2+52)(y_1^2-36)=0$

Since $y_1^2>0$, we get $y_1^2=36$, which gives $y_1=6$ (taking the positive root by symmetry).

Substituting $y_1=6$, we get $x_1=\frac{36}{12}=3$. Thus, $P$ is $(3,6)$.

For $Q$, $y_2=12$ and $x_2=\frac{144}{12}=12$. 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 $\overrightarrow{SQ}$.

$\cos \alpha =\frac{\overrightarrow{SP}\cdot \overrightarrow{SQ}}{|\overrightarrow{SP}||\overrightarrow{SQ}|}$

$\cos \alpha =\frac{0(9)+6(12)}{6\times \sqrt{9^2+12^2}}=\frac{72}{6\times 15}=\frac{72}{90}=\frac{4}{5}$

Therefore, $\sin \alpha =\sqrt{1-{\cos }^2\alpha }=\sqrt{1-{\left(\frac{4}{5}\right)}^2}=\frac{3}{5}$

Answer: $\frac{3}{5}$