Question

If $e_1$ is the eccentricity of the conic $9x^2+4y^2=36$ and $e_2$ is the eccentricity of the conic $9x^2-4y^2=36$, then

Answer 1

Calculating $e_1$
Equation of ellipse is $9x^2+4y^2=36$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1$

$\Rightarrow a^2=4,b^2=9(\text{on comparing with standard form:}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1)$

Eccentricity $e_1=\sqrt{1-\frac{a^2}{b^2}}$

$\Rightarrow e_1=\sqrt{(1-\frac{4}{9}}$

$\Rightarrow e_1=\frac{\sqrt{5}}{3}$

$\Rightarrow e_1^2=\frac{5}{9}\dots \left(1\right)$

Now, calculating $e_2$

Equation of the hyperbola is $9x^2-4y^2=36$,

$\Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1$

$\Rightarrow a^2=4,b^2=9(\text{on comparing with standard form:}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1)$

Eccentricity $e_2=\sqrt{1+\frac{b^2}{a^2}}$

$\Rightarrow e_2=\sqrt{(1+\frac{9}{4}}$

$\Rightarrow e_2=\frac{\sqrt{13}}{2}$

$\Rightarrow e_2^2=\frac{13}{4}\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get

$e_2^2-e_1^2=\frac{13}{4}-\frac{5}{9}$

$\Rightarrow e_2^2-e_1^2=\frac{117-20}{36}$

$\Rightarrow e_2^2-e_1^2=\frac{97}{36}$

$\Rightarrow e_2^2-e_1^2=2.69$

$\Rightarrow 2<e_2^2-e_1^2<3$

Hence, option (B) is correct.