Let the foci of a hypebola H coincide with the foci of the ellipse and the eccentriticy of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is and the length of its conjugate axis is , the

Question

Let the foci of a hypebola H coincide with the foci of the ellipse $E : \frac{{(x - 1)}^{2}}{100} + \frac{{(y - 1)}^{2}}{75} = 1$ and the eccentriticy of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is $α$ and the length of its conjugate axis is $β$ , then $3 α^{2} + 2 β^{2}$ is equal to [2024]

Smart Achievers · Accepted Answer

(2)

Eccentricity of ellipse E i.e., $e_{E} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$

$= \sqrt{1 - \frac{75}{100}} = \sqrt{\frac{25}{100}} = \frac{1}{2}$

So, eccentricity of hyperbola = 2

Now, foci of ellipse = (1 $\pm$ ae, 1) = (1 $\pm$ 5, 1)

= (6, 1) and (–4, 1) = Foci of hyperbola

Now, distance between foci = $2 a e_{H}$

$\Rightarrow \sqrt{10^{2} + 0} = 4 a \Rightarrow a = \frac{5}{2}$

Also, $e_{H}^{2} = 1 + \frac{b^{2}}{a^{2}}$

$\Rightarrow {(2)}^{2} = 1 + \frac{4 b^{2}}{25} \Rightarrow \frac{75}{4} = b^{2} \Rightarrow b = \frac{\sqrt{75}}{2}$

Length of transverse axis of H = 2a = $2 \times \frac{5}{2}$

$\Rightarrow α = 5$

Length of conjugate axis of H = 2b = $\sqrt{75}$

$\Rightarrow β = \sqrt{75}$

Now, $3 α^{2} + 2 β^{2} = 75 + 150 = 225$ .

Solution

1 237

2 225

3 205

4 242

Want Us to Email you About Special Offers & Updates?