Question 23. If the points of intersection of two distinct conics x^{2}+y^{2}=4b and
\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1 lie on the curve y^{2}=3x^{2},
then 3\sqrt{3} times the area of the rectangle formed by the intersection points is ____.
Answer : 432
Explanation:
Putting y2 = 3x2 in both the conics
we get x2 = b and \frac b{16}+\frac3b=1
⇒ b = 4, 12 (b = 4 is rejected because curves coincide)
∴ b = 12
Hence points of intersection are
\left(\pm\sqrt{12},\;\pm6\right) ⇒ area of rectangle = 432