Question

The normal at $P$ to the hyperbola $\frac{x^2}{9}-\frac{y^2}{1}=1$ meets the transverse axis $A{A}^{\prime }$ at $G$ and the conjugate axis $B{B}^{\prime }$ at $H.$ If $CF$ is the perpendicular drawn from centre $C$ of the hyperbola to the normal, then

• A. $PF\cdot PG=C{B}^2$
• B. $PF\cdot PG=2C{B}^2$
• C. Locus of mid-point of $GH$ is another hyperbola with eccentricity $=\sqrt{10}$
• D. $PF\cdot PH=C{A}^2$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $PF\cdot PG=C{B}^2$
C Locus of mid-point of $GH$ is another hyperbola with eccentricity $=\sqrt{10}$
D $PF\cdot PH=C{A}^2$

Given : $\frac{x^2}{9}-\frac{y^2}{1}=1$
$C=(0,0),A=(3,0),B=(0,1)$
Let $P=(3\sec \theta ,\tan \theta )$
Equation of normal in parametric form is
$\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2\Rightarrow 3x\cos \theta +y\cot \theta =10$

$\therefore G\equiv (\frac{10\sec \theta }{3},0),H\equiv (0,10\tan \theta )$

$PG=\sqrt{\frac{{\sec }^2\theta }{9}+{\tan }^2\theta }\Rightarrow PG=\frac{\sqrt{1+9{\sin }^2\theta }}{3|\cos \theta |}PH=\sqrt{9{\sec }^2\theta +81{\tan }^2\theta }\Rightarrow PH=\frac{3\sqrt{1+9{\sin }^2\theta }}{|\cos \theta |}$

Now, the equation of $CF$ is
$x\cot \theta -3y\cos \theta =0$
Distance from $P$
$PF=\frac{|\frac{3}{\sin \theta }-3\sin \theta |}{\sqrt{{\cot }^2\theta +9{\cos }^2\theta }}\Rightarrow PF=\frac{3{\cos }^2\theta }{\sqrt{{\cos }^2\theta +9{\sin }^2\theta {\cos }^2\theta }}\Rightarrow PF=\frac{3|\cos \theta |}{\sqrt{1+9{\sin }^2\theta }}PF\cdot PG=1=C{B}^{2}PF\cdot PH=9=C{A}^2$

Let the midpoint of $G$ and $H$ be $Q(h,k)$, then
$(h,k)=(\frac{5\sec \theta }{3},5\tan \theta ){\sec }^2\theta -{\tan }^2\theta =1\Rightarrow {\left(\frac{3h}{5}\right)}^2-{\left(\frac{k}{5}\right)}^2=1$

Therefore, the required locus is
$\frac{x^2}{\frac{25}{9}}-\frac{y^2}{25}=1$
This is a hyperbola, whose eccentricity
$=\sqrt{1+\frac{25}{\frac{25}{9}}}=\sqrt{10}$