JEE PYQ: Ellipse Question 1
Question 1 - 2021 (16 Mar Shift 2)
If the point of intersections of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ and the circle $x^2 + y^2 = 4b$, $b > 4$ lie on the curve $y^2 = 3x^2$, then $b$ is equal to:
(a) 12 (b) 5 (c) 6 (d) 10
Show Answer
Answer: (a) 12
Solution
$y^2 = 3x^2$ and $x^2 + y^2 = 4b$
Solve both we get $x^2 = b$
$\frac{x^2}{16} + \frac{3x^2}{b^2} = 1$
$\frac{b}{16} + \frac{3}{b} = 1$
$b^2 - 16b + 48 = 0$
$(b - 12)(b - 4) = 0$
$b = 12$, $b > 4$
BETA
