Question

Find the locus of the mid-points 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.

• A. $\frac{x^2}{4a^2}-\frac{y^2}{4b^2}=1$
• B. $(\frac{x^2}{a^2}-\frac{y^2}{b^2})=\frac{x^2}{a^4}+\frac{y^2}{b^4}$
• C. $\frac{x}{a}-\frac{y}{b}=\frac{1}{a^2}+\frac{1}{b^2}$
• D. ${(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=\frac{x^2}{a^4}+\frac{y^2}{b^4}$

Answer 1

The correct option is D ${(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=\frac{x^2}{a^4}+\frac{y^2}{b^4}$

Let $C$ be the center and $PQ$ a chord of the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ...(1)
If $(x_1,y_1)$ is the mid point of $PQ$ the its equation is given by $T=S_1$
is $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}$ ...(2)
Now the point $C$ is the origin.
Therefore the combined equation of the line $CP$ and $CQ$
joining the origin $C$ to the point of intersection of (2) and (1) is
obtained by making the equation (1) homogeneous with the help of (2)
So it is given by
$\frac{x^2}{a^2}-\frac{y^2}{b^2}={\left\{\frac{\frac{{xx}_1}{a^2}-\frac{y{y}_1}{b^2}}{\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}}\right\}}^2$
$\Rightarrow (\frac{x^2}{a^2}-\frac{y^2}{b^2}){(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2})}^2{-(\frac{{xx}_1}{a^2}-\frac{y{y}_1}{b^2})}^2=0$
Now $PQ$ subtends a right angle at the center $(0,0)$
Therefore the lines $CP$ and $CQ$ given by (3)
are at right angles.
Hence in the equation (3)
The coeff of $x^2$ = coeff. of $y^2$ =0
$\{\frac{1}{a^2}{(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2})}^2-\frac{x_1^2}{a^4}\}+\{\frac{1}{b^2}{(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2})}^2-\frac{y_1^2}{b^4}\}=0$
$\Rightarrow {\{\frac{x_1^2}{a^2}-\frac{y_1^2}{a^2}\}}^2(\frac{1}{a^2}-\frac{1}{b^2})=\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}$
Therefore the locus of $(x,y)$ is
${(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2(\frac{1}{a^2}-\frac{1}{b^2})=\frac{x^2}{a^4}+\frac{y^2}{b^4}$