Q106 / 180 · KCET 2013 · Mathematics
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
- Correct answerA2
- B$\sqrt{\frac{7}{3}}$
- C$\sqrt{7}$
- D$\sqrt{3}$
Solution
Given, equation of ellipse is,
$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$
and equation of hyperbola is
$\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$
$\therefore$ Eccentricity of ellipse is
$e_{1}=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{4-3}{4}}=\frac{1}{2}$
and eccentricity of hyperbola is
$e_{2}=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{4+3}{4}}=\frac{\sqrt{7}}{2}$
$\therefore$ Sum of the squares of the eccentricities
$\begin{aligned}
&=e_{1}^{2}+e_{2}^{2}=\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{7}}{2}\right)^{2} \\
&=\frac{1}{4}+\frac{7}{4}=\frac{8}{4}=2
\end{aligned}$
Hence, the correct answer is option: 'A'.