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

Putting y^2=3 x^2 in both the conics We get x^2=b and (b / 16)+(3 / b)=1 => b=4,12 ( b=4 is rejected because curves coincide) Therefore b=12 Hence points of intersection are ( ± sqrt(12), ± 6) => area of rectangle =432

If the points of intersection of two distinct conics x2... | null (Mathematics)

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

Explanation

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