Question

If e₁ is the eccentricity of the conic 9x² + 4y² = 36 and e₂ is the eccentricity of the conic 9x² − 4y² = 36, then

विकल्प

• e₁² − e₂² = 2

• 2 < e₂² − e₁² < 3

• e₂² − e₁² = 2

• e₂² − e₁² > 3

Answer 1

2 < e₂² − e₁² < 3
The conic ​ 

\[9 x^2 + 4 y^2 = 36\]  can rewritten in the following way:

\[\frac{9 x^2}{36} + \frac{4 y^2}{36} = 1\]

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

This is the standard equation of an ellipse.

\[\therefore b^2 = a^2 \left( 1 - e_1 \right)^2 \]

\[ \Rightarrow 9 = 4 \left( 1 - e_1 \right)^2 \]

\[ \Rightarrow \left( e_1 \right)^2 = \frac{- 5}{4}\]

The conic ​

\[9 x^2 - 4 y^2 = 36\]  can rewritten in the following way:

\[\frac{9 x^2}{36} - \frac{4 y^2}{36} = 1\]

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

This is the standard equation of a hyperbola.

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

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

\[ \Rightarrow \left( e_2 \right)^2 = \frac{13}{4}\]

\[\therefore {e_2}^2 - {e_1}^2 = \frac{13}{4} + \frac{5}{4} = 2 . 5\]