JEE Main 2025 Jan 28 Shift 1, Mathematics Q2: Parabola: Chord, Tangent and Normal

JEE Main2025Jan 28, Shift 1Mathematics

Let ABCD be a trapezium whose vertices lie on the parabola $y^{2} = 4 x$ . Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length $\frac{25}{4}$ and it passes through the point $(1, 0)$ , then the area of ABCD is :

A
$\frac{75}{4}$
B
$\frac{25}{2}$
C
$\frac{125}{8}$
D
$\frac{75}{8}$

Solution

For the parabola $y^{2} = 4 x$ we have $a = 1$ . Parameterise points on the parabola as $(t^{2}, 2 t)$ .

Since AD and BC are vertical, the trapezium is symmetric about the x-axis. Let $A = (t_{1}^{2}, 2 t_{1})$ and $D = (t_{1}^{2}, - 2 t_{1})$ on the left chord, and $B = (t_{2}^{2}, 2 t_{2})$ , $C = (t_{2}^{2}, - 2 t_{2})$ on the right chord.

The diagonal AC joins $A (t_{1}^{2}, 2 t_{1})$ to $C (t_{2}^{2}, - 2 t_{2})$ . For AC to pass through the focus $(1, 0)$ , the chord through the focus uses the focal-chord relation $t_{1} t_{2} = - 1$ , equivalently $C = (1/ t_{1}^{2}, - 2/ t_{1})$ .

Focal-chord length: $∣ A C ∣ = (t_{1} + \frac{1}{t _{1}})^{2} = \frac{25}{4}$ , giving $t_{1} + \frac{1}{t _{1}} = \frac{5}{2}$ , so $t_{1} = 2$ (taking the positive root above the axis).

Then $A = (4, 4)$ , $D = (4, - 4)$ , $B = (1/4, 1)$ , $C = (1/4, - 1)$ .

The two parallel sides have lengths $∣ A D ∣ = 8$ and $∣ B C ∣ = 2$ ; their separation is $4 - \frac{1}{4} = \frac{15}{4}$ .

Area $= \frac{1}{2} (8 + 2) \cdot \frac{15}{4} = \frac{75}{4}$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q2: Parabola: Chord, Tangent and Normal

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