KCET · Maths · Ellipse
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 \(\sqrt{\frac{7}{3}}\)
- C \(\sqrt{7}\)
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
&=e_{1}^{2}+e_{2}^{2}=\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{7}}{2}\right)^{2} \\
&=\frac{1}{4}+\frac{7}{4}=\frac{8}{4}=2
\end{aligned}\)
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
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
\(\begin{aligned}
&=e_{1}^{2}+e_{2}^{2}=\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{7}}{2}\right)^{2} \\
&=\frac{1}{4}+\frac{7}{4}=\frac{8}{4}=2
\end{aligned}\)
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