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(D) The equation of the given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Let $(h, k)$ be the pole of a chord $PQ$ of the hyperbola. The equatio

Vedclass · Accepted Answer

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

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(D) The equation of the given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Let $(h, k)$ be the pole of a chord $PQ$ of the hyperbola. The equation of the chord of contact $PQ$ is $\frac{xh}{a^2}-\frac{yk}{b^2}=1$. To find the combined equation of the lines joining the origin to the points of intersection of the hyperbola and the chord,we homogenize the hyperbola equation using the chord equation: $\frac{x^2}{a^2}-\frac{y^2}{b^2} = \left(\frac{xh}{a^2}-\frac{yk}{b^2}\right)^2$. Since the chord $PQ$ subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero. Expanding the right side: $\frac{x^2}{a^2}-\frac{y^2}{b^2} = \frac{x^2h^2}{a^4} + \frac{y^2k^2}{b^4} - \frac{2xyhk}{a^2b^2}$. Rearranging terms: $x^2\left(\frac{1}{a^2}-\frac{h^2}{a^4}\right) + y^2\left(-\frac{1}{b^2}-\frac{k^2}{b^4}\right) + \frac{2xyhk}{a^2b^2} = 0$. Setting the sum of coefficients of $x^2$ and $y^2$ to zero: $\left(\frac{1}{a^2}-\frac{h^2}{a^4}\right) + \left(-\frac{1}{b^2}-\frac{k^2}{b^4}\right) = 0$. $\Rightarrow \frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2} - \frac{1}{b^2}$. Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2} - \frac{1}{b^2}$.

The locus of a variable point whose chord of contact with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends a right angle at the origin is

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