If the eccentricity of the hyperbola times the eccentricity of the ellipse , then a value of α can be nh?kZo`Ùk dh mRdsUærk xquk gS] rc α

JEE MainMathematicsEasy / Moderate/ ToughHyperbola

If the eccentricity of the hyperbola
$x^{2} - y^{2}\sec^{2}\alpha = 5\ is\ \sqrt{3\ }$ times the eccentricity
of the ellipse $x^{2}\sec^{2}\alpha + y^{2} = 25$ , then a value of α can
be
nh?kZo`Ùk $x^{2}\sec^{2}\alpha + y^{2} = 25$ dh mRdsUærk
$\ \sqrt{3\ }$ xquk gS] rc α

Show answer and solution

For the hyperbola

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

vfrijoy;

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

ds fy,
We have;gka

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

nh?kZo`r

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

ds
fy,
We have;gka

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

Or;k

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

Or;k

2 = 4\sin^{2}\alpha

Or;k

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

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

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