For the Hyperbola \; x²/² α - \; y²/² α = 1, which of the following remains constant when α varies =? [AIEEE 2007]

Correct Answer is : abscissae of foci

Correct Answer is : abscissae of foci

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Hyperbola

Section: Mathematics

Question : 96 of 98

Marks: +1, -0

For the Hyperbola

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

\alpha

=?

Solution:

Given, equation of hyperbola is

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

We know that the equation of hyperbola is

\Rightarrow \sin^2\alpha = \cos^2\alpha (e^2-1)

\Rightarrow \sin^2\alpha + \cos^2\alpha = \cos^2\alpha \cdot e^2

\Rightarrow e^2 = 1 + \tan^2\alpha = \sec^2\alpha \Rightarrow e = \sec\alpha

\therefore a e = \cos\alpha \cdot \; \frac{1}{\cos\alpha} = 1

Co-ordinate of foci are

(\pm \alpha e, 0)

i.e.

(\pm 1,0)

Hence, abscissae of foci remain constant when

\alpha

varies.

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