Question

Find the locus of the mid point of the chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which subtend a right angle at the origin.

Answer 1

Let (h, k) be the mid-point of the chord of the hyperbola. Then its equation is $\frac{hx}{a^2}-\frac{ky}{b^2}-1=\frac{h^2}{b^2}-\frac{k^2}{b^2}-1or\frac{hx}{a^2}-\frac{ky}{b^2}=\frac{h^2}{a^2}-\frac{k^2}{b^2}......\left(1\right)$

The equation of the lines joining the origin to the points of intersection of the hyperbola and the chord (1) is obtained by making homogeneous hyperbola with the help of (1)

$\therefore \frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{{(\frac{hx}{a^2}-\frac{ky}{b^2})}^2}{{(\frac{h^2}{a^2}-\frac{k^2}{b^2})}^2}\Rightarrow \frac{1}{a^2}{(\frac{h^2}{a^2}-\frac{k^2}{b^2})}^2{x}^2-\frac{1}{b^2}{(\frac{h^2}{a^2}-\frac{k^2}{b^2})}^2{y}^2=\frac{h^2}{a^4}{x}^2+\frac{k^2}{b^4}{y}^2-\frac{2hk}{a^2{b}^2}xy.....\left(2\right)$

The lines represented by (2) will be at right angle if coefficient of $x^2+$ coefficient of $y^2=0$.

$\Rightarrow \frac{1}{a^2}{(\frac{h^2}{a^2}-\frac{k^2}{b^2})}^2-\frac{h^2}{a^2}-\frac{1}{b^2}{(\frac{h^2}{a^2}-\frac{k^2}{b^2})}^2-\frac{k^2}{b^4}=0\Rightarrow {(\frac{h^2}{a^2}-\frac{k^2}{b^2})}^2(\frac{1}{a^2}-\frac{1}{b^2})=\frac{h^2}{a^4}+\frac{k^2}{b^4}$

Hence, the locus of (h, k) is ${(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2(\frac{1}{a^2}-\frac{1}{b^2})=\frac{x^2}{a^4}+\frac{y^2}{b^4}$