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  • The locus of the poles of the chords of the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%

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

Option: 1

\mathrm{\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}}


Option: 2

\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}}


Option: 3

\mathrm{\frac{x^2}{a^4}-\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}}


Option: 4

\mathrm{\frac{x^2}{a^4}-\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}}


Answers (1)

best_answer

Let \mathrm{(x_{1},y_{1})}  be the pole w.r.t. \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\ \ \ ............(i)}
Then equation of polar is \mathrm{\frac{h x}{a^2}-\frac{k y}{b^2}=1\ \ \ ............(ii)}
The equation of lines joining the origin to the points of intersection of \mathrm{(i)} and \mathrm{(ii)} is obtained by
making homogeneous \mathrm{(i)} with the help of \mathrm{(ii)}, then \mathrm{\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)=\left(\frac{h x}{a^2}-\frac{k y}{b^2}\right)^2\Rightarrow }
\mathrm{x^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{2 h k}{a^2 b^2} x y=0}
Since the lines are perpendicular, then coefficient of \mathrm{x^2+} coefficient of \mathrm{y^2=0}
\mathrm{\frac{1}{a^2}-\frac{h^2}{a^4}-\frac{1}{b^2}-\frac{k^2}{b^4}=0 \quad \text { or } \quad \frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}} Hence required locus is
\mathrm{\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}}

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Rishabh

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