Question

If e₁ and e₂ are respectively the eccentricities of the ellipse $\frac{x^2}{18}+\frac{y^2}{4}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, then the relation between e₁ and e₂ is
(a) 3 e₁² + e₂² = 2
(b) e₁² + 2 e₂² = 3
(c) 2 e₁² + e₂² = 3
(d) e₁² + 3 e₂² = 2

Answer 1

(c) 2 e₁² + e₂² = 3
The standard form of the ellipse is $\frac{x^2}{18}+\frac{y^2}{4}=1$, where $a^2=18\mathrm{and}{b}^2=4$.
So, the eccentricity is calculated in the following way:
$$\begin{gathered} b^2=a^2(1-{e_1}^2) \\ \Rightarrow 4=18(1-{e_1}^2) \\ \Rightarrow \frac{2}{9}=1-{e_1}^2 \\ \Rightarrow {e_1}^2=\frac{7}{9} \end{gathered}$$
The standard form of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$, where $a^2=9\mathrm{and}{b}^2=4$.
So, the eccentricity is calculated in the following way:
$$\begin{gathered} b^2=a^2({e_2}^2-1) \\ \Rightarrow 4=9({e_2}^2-1) \\ \Rightarrow \frac{4}{9}={e_2}^2-1 \\ \Rightarrow {e_2}^2=\frac{13}{9} \end{gathered}$$

$$\begin{gathered} \therefore 2{e_1}^2+{e_2}^2=\frac{2\times 7}{9}+\frac{13}{9} \\ =\frac{27}{9} \\ =3 \end{gathered}$$