Maths Q12 – JEE-MAIN 2026

Let chord PQ of length

3 \sqrt{13}

of the parabola

y^{2} = 12 x

be such that the ordinates of points P and Q are in the ratio 1:2. If the chord PQ subtends an angle

α

at the focus of the parabola, then

\sin α

is equal to:

Clue 1 of 3: Parabola Properties

Identify the standard form of the parabola equation to find its focus and parameter 'a'.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Parabola Parameters and Points✦ Active

The given parabola is $y^{2} = 12 x$ . Comparing with the standard form $y^{2} = 4 a x$ , we find $4 a = 12$ , so $a = 3$ . The focus of the parabola is $S (a, 0) = (3, 0)$ . Let the ordinates of points P and Q be $y_{P} = k$ and $y_{Q} = 2 k$ respectively. Since $x = y^{2} / 12$ , the coordinates of P and Q are $P (k^{2} / 12, k)$ and $Q ((2 k)^{2} / 12, 2 k) = Q (k^{2} / 3, 2 k)$ .

Step 2: Use Chord Length to Determine 'k'○ Expand

The length of the chord PQ is given as $3 \sqrt{13}$ . Using the distance formula:

(3 \sqrt{13})^{2} = {(\frac{k^{2}}{3} - \frac{k^{2}}{12})}^{2} + (2 k - k)^{2}

\Rightarrow 117 = {(\frac{4 k^{2} - k^{2}}{12})}^{2} + k^{2}

\Rightarrow 117 = {(\frac{3 k^{2}}{12})}^{2} + k^{2}

\Rightarrow 117 = {(\frac{k^{2}}{4})}^{2} + k^{2}

\Rightarrow 117 = \frac{k^{4}}{16} + k^{2}

Let $m = k^{2}$ . The equation becomes $117 = \frac{m^{2}}{16} + m$ , which simplifies to $m^{2} + 16 m - 1872 = 0$ . Solving this quadratic equation for $m$ : $m = \frac{- 16 \pm \sqrt{16^{2} - 4 (1) (- 1872)}}{2} = \frac{- 16 \pm \sqrt{256 + 7488}}{2} = \frac{- 16 \pm \sqrt{7744}}{2} = \frac{- 16 \pm 88}{2}$ . Since $m = k^{2}$ must be positive, we take $m = \frac{- 16 + 88}{2} = \frac{72}{2} = 36$ . Thus, $k^{2} = 36$ , so $k = \pm 6$ . Choosing $k = 6$ , the points are $P (36 / 12, 6) = (3, 6)$ and $Q (36 / 3, 12) = (12, 12)$ .

Step 3: Calculate

\sin α

using Vectors○ Expand

The focus is $S (3, 0)$ . The vectors from the focus to P and Q are:

\vec{S P} = P - S = (3 - 3, 6 - 0) = (0, 6)

\vec{S Q} = Q - S = (12 - 3, 12 - 0) = (9, 12)

The angle $α$ between $\vec{S P}$ and $\vec{S Q}$ can be found using the dot product formula $\vec{S P} \cdot \vec{S Q} = | \vec{S P} | | \vec{S Q} | \cos α$ :

\vec{S P} \cdot \vec{S Q} = (0) (9) + (6) (12) = 72

| \vec{S P} | = \sqrt{0^{2} + 6^{2}} = 6

| \vec{S Q} | = \sqrt{9^{2} + 12^{2}} = \sqrt{81 + 144} = \sqrt{225} = 15

\cos α = \frac{72}{6 \times 15} = \frac{72}{90} = \frac{4}{5}

Since $α$ is an angle subtended by a chord at the focus, $0 < α < π$ , so $\sin α \geq 0$ . Using the identity $\sin^{2} α + \cos^{2} α = 1$ :

\sin α = \sqrt{1 - \cos^{2} α} = \sqrt{1 - {(\frac{4}{5})}^{2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}

💡 Teacher's Secret Hint

Remember that for an angle in a triangle, the sine value is always non-negative.

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Maths Question 12 – JEE-MAIN 2026

Clue 1 of 3: Parabola Properties

🧠 Full Solution Path