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Question: The locus of the poles of the chords of the hyperbola \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = ...

The locus of the poles of the chords of the hyperbola x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, which subtend a right angle at the centre is

A

x2a4+y2b4=1a2−1b2\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}

B

x2a2−y2b2=1a2−1b2\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}

C

x2a4−y2b4=1a2+1b2\frac{x^{2}}{a^{4}} - \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

D

x2a4−y2b4=1a2−1b2\frac{x^{2}}{a^{4}} - \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}

Answer

x2a4+y2b4=1a2−1b2\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}

Explanation

Solution

Let CSCS be the pole w.r.t. x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 ......(i)

Then equation of polar is hxa2−kyb2=1\frac{hx}{a^{2}} - \frac{ky}{b^{2}} = 1 .....(ii)

The equation of lines joining the origin to the points of intersection of (i) and (ii) is obtained by making homogeneous (i) with the help of (ii), then

(x2a2−y2b2)=(hxa2−kyb2)2\left( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} \right) = \left( \frac{hx}{a^{2}} - \frac{ky}{b^{2}} \right)^{2}

⇒ x2(1a2−h2a4)−y2(1b2+k2b4)+2hka2b2xy=0x^{2}\left( \frac{1}{a^{2}} - \frac{h^{2}}{a^{4}} \right) - y^{2}\left( \frac{1}{b^{2}} + \frac{k^{2}}{b^{4}} \right) + \frac{2hk}{a^{2}b^{2}}xy = 0Since the lines are perpendicular, then coefficient of x2+x^{2} + coefficient of y2=0y^{2} = 0

1a2−h2a4−1b2−k2b4=0\frac{1}{a^{2}} - \frac{h^{2}}{a^{4}} - \frac{1}{b^{2}} - \frac{k^{2}}{b^{4}} = 0 or h2a4+k2b4=1a2−1b2\frac{h^{2}}{a^{4}} + \frac{k^{2}}{b^{4}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}.

Hence required locus is x2a4+y2b4=1a2−1b2\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}