# If the Eccentricity of the Hyperbola X2 − Y2 Sec2α = 5 is √ 3 Times the Eccentricity of the Ellipse X2 Sec2 α + Y2 = 25, Then α = - Mathematics

MCQ

If the eccentricity of the hyperbola x2 − y2 sec2α = 5 is $\sqrt{3}$  times the eccentricity of the ellipse x2 sec2 α + y2 = 25, then α =

#### Options

• $\frac{\pi}{6}$

• $\frac{\pi}{4}$

• $\frac{\pi}{3}$

• $\frac{\pi}{2}$

#### Solution

$\frac{\pi}{4}$

The hyperbola

$x^2 - y^2 se c^2 \alpha = 5$ can be rewritten in the following way:

$\frac{x^2}{5} - \frac{y^2}{5 \cos^2 \alpha} = 1$

This is the standard form of a hyperbola, where

$a^2 = 5 \text { and } b^2 = 5 \cos^2 \alpha$.

$\Rightarrow b^2 = a^2 \left( {e_1}^2 - 1 \right)$

$\Rightarrow 5 \cos^2 \alpha = 5\left( {e_1}^2 - 1 \right)$

$\Rightarrow {e_1}^2 = \cos^2 \alpha + 1 . . . . . \left( 1 \right)$

The ellipse

$x^2 se c^2 \alpha + y^2 = 25$ can be rewritten in the following way:

$\frac{x^2}{25 \cos^2 \alpha} + \frac{y^2}{25} = 1$

This is the standard form of an ellipse, where

$a^2 = 25 \text { and } b^2 = 25 \cos^2 \alpha$.

$b^2 = a^2 \left( 1 - {e_2}^2 \right)$

$\Rightarrow {e_2}^2 = 1 - \cos^2 \alpha$

$\Rightarrow {e_2}^2 = \sin^2 \alpha . . . . . . (2)$

According to the question,

$\cos^2 \alpha + 1 = 3\left( \sin^2 \alpha \right)$

$\Rightarrow 2 = 4 \sin^2 \alpha$

$\Rightarrow \sin\alpha = \frac{1}{\sqrt{2}}$

$\Rightarrow \alpha = \frac{\pi}{4}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Q 13 | Page 19