JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 26)
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 _________.
Answer
432
Explanation
Putting $$y^2=3 x^2$$ in both the conics
We get $$x^2=b$$ and $$\frac{b}{16}+\frac{3}{b}=1$$
$$\Rightarrow \mathrm{b}=4,12 \quad(\mathrm{b}=4$$ is rejected because curves coincide)
$$\therefore \mathrm{b}=12$$
Hence points of intersection are
$$( \pm \sqrt{12}, \pm 6) \Rightarrow \text { area of rectangle }=432$$
Comments (0)
