If the points of intersection of two distinct conics x 2 +y 2 =4 b and frac x 2 16 +frac y 2 b 2 =1 lie on the curve y 2 =3 x 2 , then 3 √(3) times the area of the rectangle formed by the intersection points is

Mathematics · Ellipse

JEE Main 2024 — 29 January, Shift 1 — Question 24

If the points of intersection of two distinct conics $x^{2}+y^{2}=4 b$ and $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ lie on the curve $y^{2}=3 x^{2}$ , then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is $\qquad$

Answer: 432

Numerical answer — enter this value.

Step-by-step solution

Sol. Putting $\mathrm{y}^{2}=3 \mathrm{x}^{2}$ in both the conics

We get $x^{2}=b$ and $\frac{b}{16}+\frac{3}{b}=1$

$\Rightarrow b=4,12 \quad(b=4$ is rejected because curves coincide)

$\therefore \mathrm{b}=12$

Hence points of intersection are

$( \pm \sqrt{12}, \pm 6)$

$\Rightarrow$ area of rectangle $=432$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse