Let a and b be positive real numbers such that a > 1 and b < a . Let P be a point in the first…

Tangent at P

$\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$ Passes through $\left(1,0\right)$

$\sec \theta =a$

Now slope of $\mathrm{AP}=1$

$\frac{b \sec \theta }{a \tan \theta }=1\Rightarrow b=\tan \theta$

Now $b^2=a^2\left(e^2-1\right)$

${\tan }^2\theta ={\sec }^2\theta \left(e^2-1\right)\Rightarrow e^2-1={\sin }^2\theta$

$e^2-1\in \left[0,1\right] \because \theta \in \left(0,\frac{\pi }{2}\right)$

$e^2\in \left[1,2\right]\Rightarrow 1<e<\sqrt{2}$

Now $\mathrm{A}\left(1,0\right)$ $P\left(asec\theta , b\tan \theta \right)=\left(se{c}^2\theta ,{\tan }^2\theta \right)$

$AP=\sqrt{{\tan }^4\theta +{\tan }^4\theta }\Rightarrow AP=\sqrt{2}{\tan }^2\theta$

$\because ∆APB$ is isosceles-right angled triangle,

So, area$=\frac{1}{2}{\left(AP\right)}^2=\frac{1}{2}\times 2{\tan }^4\theta =b^4$.