If $e_{1}$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, ab$ and $e_{2}$ is the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then $e_{1}^{2}+e_{2}^{2}=$

Question

If $e_{1}$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ and $e_{2}$ is the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then $e_{1}^{2}+e_{2}^{2}=$

Aicharya · Accepted Answer

$e_{1}=\sqrt{1-\frac{b^{2}}{a^{2}}} \quad e_{2}=\sqrt{1+\frac{b^{2}}{a^{2}}}$ $e_{1}^{2}+e_{2}^{2}=2$

If $e_{1}$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ and $e_{2}$…

Solution