JEE Main 2026 Apr 5 Shift 2, Mathematics Q12: Parabola: Chord, Tangent and Normal

JEE Main2026Apr 5, Shift 2Mathematics

Let the directrix of the parabola $P : y^{2} = 8 x$ cut the $x$ -axis at the point $A$ . Let $B (α, β)$ , $α > 1$ , be a point on $P$ such that the slope of $A B$ is $3/5$ . If $B C$ is a focal chord of $P$ , then six times the area of $△ A B C$ is:

A
80
B
160
C
174
D
192

Solution

For $y^{2} = 8 x$ , the directrix is $x = - 2$ , so $A = (- 2, 0)$ . Parametrise points on $P$ as $(2 t^{2}, 4 t)$ ; let $B = (2 t^{2}, 4 t)$ .

Slope $A B = \frac{4 t}{2 t ^{2} + 2} = \frac{3}{5} \Rightarrow 20 t = 6 t^{2} + 6 \Rightarrow 3 t^{2} - 10 t + 3 = 0$ .

Roots: $t = 3$ or $t = 1/3$ . With $α > 1$ we need $2 t^{2} > 1$ , giving $t = 3$ , so $B = (18, 12)$ .

For a focal chord, parameters satisfy $t_{1} t_{2} = - 1$ . With $t_{1} = 3$ , the other endpoint has $t_{2} = - 1/3$ , giving $C = (2/9, - 4/3)$ .

Area of $△ A B C = \frac{1}{2} det - 201 18121 2/9 - 4/31 = \frac{80}{3}$ .

$∴ 6 \times Area = 160$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q12: Parabola: Chord, Tangent and Normal

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