JEE Main 2026 — Mathematics Coordinate Geometry

2026mcqmedium

Let the directrix of the parabola $P: y^2 = 8x$ , cut $x$ -axis at the point $A$ . Let $B(\alpha, \beta)$ , $\alpha > 1$ , be a point on $P$ such that the slope of $AB$ is $3/5$ . If $BC$ is a focal chord of $P$ , then six times the area of $\triangle ABC$ is :

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$80$
B
$160$
C
$174$
D
$192$

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Solution

For the parabola $P: y^2 = 8x$ , we have $4a = 8 \Rightarrow a = 2$ .

The equation of the directrix is $x = -a \Rightarrow x = -2$ . Since the directrix cuts the $x$ -axis at $A$ , the coordinates of $A$ are $(-2, 0)$ .

Let the coordinates of point $B$ on the parabola be $(2t_1^2, 4t_1)$ .

The slope of $AB$ is given as $\dfrac{3}{5}$ .

$\dfrac{4t_1 - 0}{2t_1^2 - (-2)} = \dfrac{3}{5}$

$\dfrac{2t_1}{t_1^2 + 1} = \dfrac{3}{5}$

$10t_1 = 3t_1^2 + 3 \Rightarrow 3t_1^2 - 10t_1 + 3 = 0$

$(3t_1 - 1)(t_1 - 3) = 0 \Rightarrow t_1 = \dfrac{1}{3}$ or $t_1 = 3$ .

For $t_1 = \dfrac{1}{3}$ , $\alpha = 2\left(\dfrac{1}{3}\right)^2 = \dfrac{2}{9}$ , which is rejected since $\alpha > 1$ .

For $t_1 = 3$ , $\alpha = 2(3)^2 = 18 > 1$ . Thus, $B$ is $(18, 12)$ .

Since $BC$ is a focal chord, the parameter for $C$ is $t_2 = -\dfrac{1}{t_1} = -\dfrac{1}{3}$ .

The coordinates of $C$ are $\left(2\left(-\dfrac{1}{3}\right)^2, 4\left(-\dfrac{1}{3}\right)\right) = \left(\dfrac{2}{9}, -\dfrac{4}{3}\right)$ .

The area of $\triangle ABC$ is given by:

$\Delta = \dfrac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$

$\Delta = \dfrac{1}{2} \left| -2\left(12 - \left(-\dfrac{4}{3}\right)\right) + 18\left(-\dfrac{4}{3} - 0\right) + \dfrac{2}{9}(0 - 12) \right|$

$\Delta = \dfrac{1}{2} \left| -2\left(\dfrac{40}{3}\right) - 24 - \dfrac{24}{9} \right|$

$\Delta = \dfrac{1}{2} \left| -\dfrac{80}{3} - 24 - \dfrac{8}{3} \right| = \dfrac{1}{2} \left| -\dfrac{88}{3} - \dfrac{72}{3} \right| = \dfrac{1}{2} \left| -\dfrac{160}{3} \right| = \dfrac{80}{3}$

Six times the area of $\triangle ABC = 6 \times \dfrac{80}{3} = 160$ .

Answer: $160$

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Solution

For the parabola $P: y^2 = 8x$ , we have $4a = 8 \Rightarrow a = 2$ .

The equation of the directrix is $x = -a \Rightarrow x = -2$ . Since the directrix cuts the $x$ -axis at $A$ , the coordinates of $A$ are $(-2, 0)$ .

Let the coordinates of point $B$ on the parabola be $(2t_1^2, 4t_1)$ .

The slope of $AB$ is given as $\dfrac{3}{5}$ .

$\dfrac{4t_1 - 0}{2t_1^2 - (-2)} = \dfrac{3}{5}$

$\dfrac{2t_1}{t_1^2 + 1} = \dfrac{3}{5}$

$10t_1 = 3t_1^2 + 3 \Rightarrow 3t_1^2 - 10t_1 + 3 = 0$

$(3t_1 - 1)(t_1 - 3) = 0 \Rightarrow t_1 = \dfrac{1}{3}$ or $t_1 = 3$ .

For $t_1 = \dfrac{1}{3}$ , $\alpha = 2\left(\dfrac{1}{3}\right)^2 = \dfrac{2}{9}$ , which is rejected since $\alpha > 1$ .

For $t_1 = 3$ , $\alpha = 2(3)^2 = 18 > 1$ . Thus, $B$ is $(18, 12)$ .

Since $BC$ is a focal chord, the parameter for $C$ is $t_2 = -\dfrac{1}{t_1} = -\dfrac{1}{3}$ .

The coordinates of $C$ are $\left(2\left(-\dfrac{1}{3}\right)^2, 4\left(-\dfrac{1}{3}\right)\right) = \left(\dfrac{2}{9}, -\dfrac{4}{3}\right)$ .

The area of $\triangle ABC$ is given by:

$\Delta = \dfrac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$

$\Delta = \dfrac{1}{2} \left| -2\left(12 - \left(-\dfrac{4}{3}\right)\right) + 18\left(-\dfrac{4}{3} - 0\right) + \dfrac{2}{9}(0 - 12) \right|$

$\Delta = \dfrac{1}{2} \left| -2\left(\dfrac{40}{3}\right) - 24 - \dfrac{24}{9} \right|$

$\Delta = \dfrac{1}{2} \left| -\dfrac{80}{3} - 24 - \dfrac{8}{3} \right| = \dfrac{1}{2} \left| -\dfrac{88}{3} - \dfrac{72}{3} \right| = \dfrac{1}{2} \left| -\dfrac{160}{3} \right| = \dfrac{80}{3}$

Six times the area of $\triangle ABC = 6 \times \dfrac{80}{3} = 160$ .

Answer: $160$

JEE Main 2026 — Mathematics Coordinate Geometry

Options

Did you get this right?

Solution

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JEE Main Coordinate Geometry in other years

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Other Mathematics topics for JEE Main

Algebra

Calculus

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Keep practicing Coordinate Geometry

JEE Main 2026 — Mathematics Coordinate Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Coordinate Geometry in other years

More JEE Main Coordinate Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Vectors & 3D Geometry

Probability & Statistics

Trigonometry

Keep practicing Coordinate Geometry