JEE Main 2025 Apr 4 Shift 2, Mathematics Q14: Ellipse: Chord, Tangent and Normal

JEE Main2025Apr 4, Shift 2Mathematics

Let for two distinct values of $p$ the lines $y = x + p$ touch the ellipse $E : \frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1$ at the points $A$ and $B$ . Let the line $y = x$ intersect $E$ at the points $C$ and $D$ . Then the area of the quadrilateral $A B C D$ is equal to

A
36
B
24
C
48
D
20

Solution

For the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ ( $a = 4, b = 3$ ), the tangent $y = m x + c$ touches when $c^{2} = a^{2} m^{2} + b^{2}$ . With $m = 1$ : $c^{2} = 16 + 9 = 25$ , so $c = \pm 5$ , giving $p = \pm 5$ .

The points of contact are $(\mp \frac{a ^{2} m}{c}, \pm \frac{b ^{2}}{c}) = (\mp \frac{16}{5}, \pm \frac{9}{5})$ :

$A = (- \frac{16}{5}, \frac{9}{5}), B = (\frac{16}{5}, - \frac{9}{5})$ .

The line $y = x$ meets the ellipse where $\frac{x ^{2}}{16} + \frac{x ^{2}}{9} = 1 \Rightarrow x^{2} = \frac{144}{25} \Rightarrow x = \pm \frac{12}{5}$ .

$C = (\frac{12}{5}, \frac{12}{5}), D = (- \frac{12}{5}, - \frac{12}{5})$ .

$[△ A B D] = \frac{1}{2} ∣ x_{A} (y_{B} - y_{D}) + x_{B} (y_{D} - y_{A}) + x_{D} (y_{A} - y_{B}) ∣$ .

Substituting the coordinates yields $[△ A B D] = 12$ . By symmetry $[△ A B C] = 12$ as well, so the quadrilateral has area $24$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q14: Ellipse: Chord, Tangent and Normal

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