If the normal at P to the hyperbola $frac{x^2}{a^2}-frac{y^2}{b^2}=1$ meets the transverse axis in G and the conjugate axis in g and CF be perpendicular to the normal from the centre, then $mathrm{PF} cdot mathrm{PG}+mathrm{PF} cdot mathrm{Pg}=$

If the normal at P to the hyperbola <img alt="\frac{x^2}{a^2}-\frac{y^2}{b^2}=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1

If the normal at P to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the transverse axis in G and the conjugate axis in $\mathrm{g}$ and $\mathrm{\mathrm{CF}}$ be perpendicular to the normal from the centre, then $\mathrm{\mathrm{PF} \cdot \mathrm{PG}+\mathrm{PF} \cdot \mathrm{Pg}=}$
Option: 1
$a^2-b^2$

Option: 2
$b^2-a^2$

Option: 3
$a^2+b^2$

Option: 4
$2\left(a^2+b^2\right)$

Answers (1)

Let $\mathrm{P(a \sec \phi, b \tan \phi) }$ be any point on the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 }$ then equation of tangent and normal at $\mathrm{\mathrm{P} }$ are
$\mathrm{\frac{\mathrm{x}}{\mathrm{a}} \sec \phi-\frac{\mathrm{y}}{\mathrm{b}} \tan \phi=1 }$ ..............(1)
and $\mathrm{\quad a x \cos \phi+b y \cot \phi=a^2+b^2 }$ ............(2)
Putting y=0 & x=0 in (2), we get the points G and $\mathrm{\mathrm{g} }$ as
$\mathrm{\begin{aligned} & \left(\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}} \sec \phi, 0\right),\left(0, \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}} \tan \phi\right) \\ \therefore \quad(\mathrm{PG})^2 & =\left(\mathrm{a} \sec \phi-\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}} \sec \phi\right)^2+(\mathrm{b} \tan \phi-0)^2 \\ = & \frac{\mathrm{b}^4}{\mathrm{a}^2} \sec ^2 \phi+\mathrm{b}^2 \tan ^2 \phi \\ = & \frac{\mathrm{b}^2}{\mathrm{a}^2}\left(\mathrm{~b}^2 \sec ^2 \phi+\mathrm{a}^2 \tan ^2 \phi\right) \end{aligned} }$ .....(3)

and
$\mathrm{\begin{aligned} & (\operatorname{Pg})^2=(\mathrm{a} \sec \phi-0)^2+\left(\mathrm{b} \tan \phi-\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}} \tan \phi\right)^2 \\ & =\mathrm{a}^2 \sec ^2 \phi+\frac{\mathrm{a}^4}{\mathrm{~b}^2} \tan ^2 \phi \\ & =\frac{\mathrm{a}^2}{\mathrm{~b}^2}\left(\mathrm{~b}^2 \sec ^2 \phi+\mathrm{a}^2 \tan ^2 \phi\right) \end{aligned} }$ .......(4)
Now $\mathrm{ \mathrm{PF} }$ is clearly equal to $\mathrm{\mathrm{CL} }$ , i.e.., perpendicular from $\mathrm{\mathrm{C}(0,0) }$ on (1);
$\mathrm{\begin{aligned} & \mathrm{PF}=\mathrm{CL}=\frac{1}{\sqrt{\left(\frac{\sec ^2 \phi}{\mathrm{a}^2}+\frac{\tan ^2 \phi}{\mathrm{b}^2}\right)}} \\ & =\frac{a b}{\sqrt{\left(b^2 \sec ^2 \phi+a^2 \tan ^2 \phi\right)}}=\frac{a b}{\left(\frac{a}{b}\right) P G} \\ & \text { (from (3)) } \\ & \therefore \quad \text { PG.PF }=b^2 \\ & \mathrm{PF}=\frac{\mathrm{ab}}{\sqrt{\left(\mathrm{b}^2 \sec ^2 \phi+\mathrm{a}^2 \tan ^2 \phi\right)}}=\frac{\mathrm{ab}}{\frac{\mathrm{b}}{\mathrm{a}} \cdot \mathrm{Pg}} \\ & \text { (from 4) } \\ & \mathrm{Pg} \cdot \mathrm{PF}=\mathrm{a}^2 \\ & \end{aligned} }$