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 write the value of 2 e₁² + e₂².

Answer 1

The standard form of the ellipse is \[\frac{x^2}{18} + \frac{y^2}{4} = 1\]  where  \[a^2 = 18 \text { and } b^2 = 4\]

So, the eccentricity is calculated in the following way:

\[\Rightarrow b^2 = a^2 (1 - {e_1}^2 )\]

\[ \Rightarrow 4 = 18(1 - {e_1}^2 )\]

\[ \Rightarrow \frac{2}{9} = 1 - {e_1}^2 \]

\[ \Rightarrow {e_1}^2 = \frac{7}{9}\]

The standard form of the hyperbola is \[\frac{x^2}{9} - \frac{y^2}{4} = 1\], where \[a^2 = 9\text {  and } b^2 = 4\].

So, the eccentricity is calculated in the following way:

\[b^2 = a^2 ( {e_2}^2 - 1)\]

\[ \Rightarrow 4 = 9( {e_2}^2 - 1)\]

\[ \Rightarrow \frac{4}{9} = {e_2}^2 - 1\]

\[ \Rightarrow {e_2}^2 = \frac{13}{9}\]

\[\therefore 2 {e_1}^2 + {e_2}^2 = \frac{2 \times 7}{9} + \frac{13}{9}\]

\[ = \frac{27}{9}\]

\[ = 3\]