Question

$e_1$ and $e_2$ are respectively the eccentricities of a hyperbola and its conjugate then $\frac{1}{e_1^2}$+$\frac{1}{e_2^2}$=1.

• A. True
• B. False

Answer 1

The correct option is A True

Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $.......\left(1\right)$

And

$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ $.........\left(2\right)$ are two hyperbola conjugate to each other.

Also let, $e_1$ and $e_2$ are the eccentricities of $\left(1\right)$ and $\left(2\right)$ respectively.

Then,

$e_1^2=1+\frac{b^2}{a^2}$ and $e_2^2=1+\frac{a^2}{b^2}$

Therefore,

$\Rightarrow \frac{1}{e_1^2}+\frac{1}{e_2^2}$

$\Rightarrow \frac{1}{1+\frac{b^2}{a^2}}+\frac{1}{1+\frac{a^2}{b^2}}$

$\Rightarrow \frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}$

$\Rightarrow \frac{a^2+b^2}{a^2+b^2}$

$\Rightarrow 1$

Hence, proved.