Question

If e₁ is the eccentricity of the conic 9x² + 4y² = 36 and e₂ is the eccentricity of the conic 9x² − 4y² = 36, then
(a) e₁² − e₂² = 2
(b) 2 < e₂² − e₁² < 3
(c) e₂² − e₁² = 2
(d) e₂² − e₁² > 3

Answer 1

(b) 2 < e₂² − e₁² < 3
The conic ​$9x^2+4y^2=36$ can rewritten in the following way:
$$\begin{gathered} \frac{9x^2}{36}+\frac{4y^2}{36}=1 \\ \Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1 \end{gathered}$$
This is the standard equation of an ellipse.
$$\begin{gathered} \therefore b^2=a^2{\left(1-e_1\right)}^2 \\ \Rightarrow 9=4{\left(1-e_1\right)}^2 \\ \Rightarrow {\left(e_1\right)}^2=\frac{-5}{4} \end{gathered}$$
The conic ​$9x^2-4y^2=36$ can rewritten in the following way:
$$\begin{gathered} \frac{9x^2}{36}-\frac{4y^2}{36}=1 \\ \Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1 \end{gathered}$$
This is the standard equation of a hyperbola.
$$\begin{gathered} \therefore b^2=a^2\left({e_2}^2-1\right) \\ \Rightarrow 9=4\left({e_2}^2-1\right) \\ \Rightarrow {\left(e_2\right)}^2=\frac{13}{4} \end{gathered}$$

$\therefore {e_2}^2-{e_1}^2=\frac{13}{4}+\frac{5}{4}=2.5$