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Question: If a variable straight line \(x\cos\alpha + y\sin\alpha = p\), which is a chord of the hyperbola \(\...

If a variable straight line xcos⁡α+ysin⁡α=px\cos\alpha + y\sin\alpha = p, which is a chord of the hyperbola x2a2−y2b2=1(b>a)\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1(b > a), subtend a right angle at the centre of the hyperbola then it always touches a fixed circle whose radius is

A

abb−2a\frac{ab}{\sqrt{b - 2a}}

B

aa−b\frac{a}{\sqrt{a - b}}

C

abb2−a2\frac{ab}{\sqrt{b^{2} - a^{2}}}

D

abb(b+a)\frac{ab}{b\sqrt{(b + a)}}

Answer

abb2−a2\frac{ab}{\sqrt{b^{2} - a^{2}}}

Explanation

Solution

Since xcos⁡α+ysin⁡α=p\mathbf{x}\mathbf{\cos}\mathbf{\alpha}\mathbf{+ y}\mathbf{\sin}\mathbf{\alpha}\mathbf{= p} subtends a right angle at centre i.e. (0,0). Making homogeneous equation of hyperbola x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 with the help of xcos⁡α+ysin⁡α=p\mathbf{x}\mathbf{\cos}\mathbf{\alpha}\mathbf{+ y}\mathbf{\sin}\mathbf{\alpha}\mathbf{= p} and putting coefficient of x2+x^{2} + coefficient of y2=0y^{2} = 0. We get 1a2−1b2=1p2\frac{1}{a^{2}} - \frac{1}{b^{2}} = \frac{1}{p^{2}}⇒ p=abb2−a2p = \frac{ab}{\sqrt{b^{2} - a^{2}}}

p is also the length of perpendicular drawn from (0, 0) to the line xcos⁡α+ysin⁡α=p\mathbf{x}\mathbf{\cos}\mathbf{\alpha}\mathbf{+ y}\mathbf{\sin}\mathbf{\alpha}\mathbf{= p}.

⇒ then radius of the circle = p = abb2−a2\frac{ab}{\sqrt{b^{2} - a^{2}}}