JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q27

MathematicsInteger+4 / -1Question 27 of 90

For the hyperbola $H:x^2-y^2=1$ and the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ , $a>b>0$ , let the

(1) eccentricity of E be reciprocal of the eccentricity of H, and

(2) the line $y=\sqrt{\dfrac52}x+K$ be a common tangent of E and H.

Then $4\left(a^2+b^2\right)$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 3

Eccentricities. $e_H=\sqrt2$ and $e_E=\frac1{\sqrt2}$ , so $b^2=\frac{a^2}2$ .

Common tangent with $m^2=\frac52$ . Hyperbola: $K^2=m^2-1=\frac32$ . Ellipse: $K^2=a^2m^2+b^2=3a^2$ . So $a^2=\frac12$ , $b^2=\frac14$ .

Answer: 3

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 27