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QJEEM24Ma4189: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4189 Solution: If the points of intersection of two distinct conics and lie on …

If the points of intersection of two distinct conics x2+y2=4bx^2+y^2=4 b and x216+y2b2=1\frac{x^2}{16}+\frac{y^2}{b^2}=1 lie on the curve y2=3x2y^2=3 x^2 , then 333 \sqrt{3} times the area of the rectangle formed by the intersection points is _________.

Step-by-Step Verified Solution

Putting y2=3x2y^2=3 x^2 in both the conics

We get x2=bx^2=b and b16+3b=1\frac{b}{16}+\frac{3}{b}=1

⇒b=4,12(b=4\Rightarrow \mathrm{b}=4,12 \quad(\mathrm{b}=4 is rejected because curves coincide)

∴b=12\therefore \mathrm{b}=12

Hence points of intersection are

(±12,±6)⇒ area of rectangle =432( \pm \sqrt{12}, \pm 6) \Rightarrow \text { area of rectangle }=432

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