Question

A line drawn through the point $P(-1,2)$ meets the hyperbola $xy=c^2$ at the points $A$ and $B$. (points $A$ and $B$ lie on same side of $P$) and $Q$ is a point on $AB$ such that $PA,PQ$ and $PB$ are in $H.P$ then locus of $Q$ is

A. $x-2y=2c^2$

B. $2x-y=2c^2$

C. $x+2y=2c^2$

D. $2x-y+2c^2=0$

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Given:

A line drawn through the point $P(-1,2)$ meets the hyperbola $xy=c^2$ at the points $A$ and $B$.

$P$ is same side of the points $A$ and $B$.

$Q$ is a point on $AB$.

It is given that, $PA,PQ$ and $PB$ are in H.P.

$\Rightarrow PQ=\frac{2PA.PB}{PQ+PB}$ ---(1)

Lets identify the values of $PA,PQ$ and $PB$.

Lets form required line,

Let the required equation of the line formed an angle $\theta$, then the parametric form of equation of the line is,

$\frac{x-x_1}{r\cos \theta }=\frac{y-y_1}{r\sin \theta }=1$

i.e., $x-x_1=r\cos \theta ,y-y_1=r\sin \theta$

Here, the point $P=(-1,2)$

$\Rightarrow x-(-1)=r\cos \theta$

$\Rightarrow x+1=r\cos \theta$ and $x=-1+r\cos \theta$

Since, in figure radius,$PQ=r$.

i.e., in figure, $PQ\cos \theta =x+1$ ---(i)

Similarly,

$y-2=r\sin \theta$ and $y=2+r\sin \theta$

i.e., in figure

$\Rightarrow y-2=PQ\sin \theta$ ---(ii)

Now, the coordinates $x=-1+r\cos \theta$ and $y=2+r\sin \theta$ lie on the hyperbola,

$xy=c^2$.

$\Rightarrow (-1+r\cos \theta )(2+r\sin \theta )=c^2$

$\Rightarrow -2-r\sin \theta +2r\cos \theta +r^2\cos \theta \sin \theta -c^2=0$

$\Rightarrow r^2\sin \theta \cos \theta +r(2\cos \theta -\sin \theta )-(2+c^2)=0$

Let, $r_1$ and $r_2$ are the roots of the equation.

$\Rightarrow r_1+r_2=\frac{\sin \theta -2\cos \theta }{\sin \theta \cos \theta }$ ---(iii)

$r_1.r_2=\frac{-(2+c^2)}{\sin \theta \cos \theta }$----(iv)

In figure,

$PA=r_1$ and $PB=r_2$.

Now from (1),

$\Rightarrow PQ=\frac{2PA.PB}{PQ+PB}$

$\Rightarrow PQ=\frac{2r_1.r_2}{r_1+r_2}$

$\Rightarrow PQ=\frac{2\left(\frac{-(2+c^2)}{\sin \theta \cos \theta }\right)}{\frac{\sin \theta -2\cos \theta }{\sin \theta \cos \theta }}$

$\Rightarrow PQ=\frac{-4-2c^2}{\sin \theta -2\cos \theta }$

$\Rightarrow PQ\sin \theta -2PQ\cos \theta =-4-2c^2$

$\Rightarrow y-2-2(x+1)=-4-2c^2$ [Since, $PQ\cos \theta =x+1,PQ\sin \theta =y-2$]

$\Rightarrow y-2-2x-2=-4-2c^2$

$\Rightarrow y-2-2x-2+4=-2c^2$

$\Rightarrow y-2x=-2c^2$

$\Rightarrow 2x-y=2c^2$

Therefore, the Locus of $Q$ is $2x-y=2c^2$.

Hence, Option B is correct.