Question

Find tha locus of the mid-point of the chords of the hyperbola $\frac{x^2}{2}-\frac{y^2}{3}=1$ which subtends a right angle at the origin

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let (h,k) be the mid-point of the chord of the hyperbola.

Then its equation is $T=s_1$

$\frac{hx}{2}-\frac{ky}{3}-1=\frac{h^2}{2}-\frac{k^2}{3}-1$

or

$\frac{hx}{2}-\frac{ky}{3}=\frac{h^2}{2}-\frac{k^2}{3}$ ..............(1)

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 with the help of equation of chord.

$\therefore \frac{x^2}{2}-\frac{y^2}{3}=\frac{{(\frac{hx}{2}-\frac{ky}{3})}^2}{{(\frac{h^2}{2}-\frac{k^2}{3})}^2}$

$\Rightarrow$ $\frac{1}{2}{(\frac{h^2}{2}-\frac{k^2}{3})}^2{x}^2-\frac{1}{3}{(\frac{h^2}{2}-\frac{k^2}{3})}^2{y}^2$

$=\frac{h^2}{4}{x}^2+\frac{k^2}{9}{y}^2-\frac{2hk}{2\times 3}xy$

$\Rightarrow [\frac{1}{2}{(\frac{h^2}{2}-\frac{k^2}{3})}^2-\frac{h^2}{4}]x^2-[\frac{1}{3}{(\frac{h^2}{2}-\frac{k^2}{3})}^2+\frac{k^2}{9}]y^2+\frac{hk}{3}xy=0$ ....................(2)

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

$\Rightarrow \frac{1}{2}{(\frac{h^2}{2}-\frac{k^2}{3})}^2-\frac{h^2}{4}-\frac{1}{3}{(\frac{h^2}{2}-\frac{k^2}{3})}^2+\frac{k^2}{9}=0$

${(\frac{h^2}{2}-\frac{k^2}{3})}^2[\frac{1}{2}-\frac{1}{3}]=\frac{h^2}{4}+\frac{k^2}{9}$

Locus of point (h,k) replacing h by x $\&$ k by y

We get,

$\frac{1}{6}{(\frac{x^2}{2}-\frac{y^2}{3})}^2=\frac{x^2}{4}+\frac{y^2}{9}$