Question

The sum of the squares of the eccentricities of the conics $\frac{x^2}{4}+\frac{y^2}{3}=1$ and $\frac{x^2}{4}-\frac{y^2}{3}=1$ is _____.

• A. $2$
• B. $\frac{7}{3}$
• C. $7$
• D. $3$

Answer 1

The correct option is A $2$

Eccentricity of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ is given by,

$e_1=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$

And eccentricity of the hyperbola $\frac{x^2}{4}-\frac{y^2}{3}=1$ is

$e_2=\sqrt{1+\frac{3}{4}}=\frac{\sqrt{7}}{2}$

Hence $e_1^2+e_2^2=\frac{1}{4}+\frac{7}{4}=\frac{8}{4}=2$