If the points of intersection of two distinct conics x^2+y^2=4 b and [(x^2)/(16)+(y^2)/(b^…

Question

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

Accepted Answer

Putting ${y}^{2}=3{x}^{2}$ in both the conics

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

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

$	herefore b=12$

Hence points of intersection are

$(\pm \sqrt{12},\pm 6)\Rightarrow =432$

Solutions · 1