If the eccentricity of the hyperbola times the eccentricity of the ellipse x^2 sec^2α + y^2 = 25, then a value of α can be.

JEE MainMathematicsModerateConic Sections

If the eccentricity of the hyperbola
$x^{2} - y^{2}\sec^{2}\alpha = 5\ is\ \sqrt{3}$ times the eccentricity
of the ellipse x^2 sec^2α + y^2 = 25, then a value of α can be.

Show answer and solution

For the
hyperbola

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

We have

e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{5\cos^{2}\alpha}{5} = 1 + \cos^{2}\alpha

For the ellipse

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

We have

e_{2}^{2} = 1 - \frac{25\cos^{2}\alpha}{25} = \sin^{2}\alpha

Given that

e_{1} = \sqrt{3}e_{2}\therefore e_{1}^{2} = 3e_{2}^{2}

1 + \cos^{2}\alpha = 3\sin^{2}\alpha

2 = 4\sin^{2}\alpha

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

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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