Let a and b be positive real numbers such that a > 1 and b < a ....

Tangent at P

$\dfrac{xsec\theta}{a} - \dfrac{ytan\theta}{b} = 1$ Passes through \left(1, 0\right)$</p><p>$sec\theta = a$</p><p>Now slope of$AP = 1$</p><p>$\dfrac{bsec\theta}{atan\theta} = 1 ⇒ b = tan\theta$</p><p>Now${b}^{2} = {a}^{2}\left({e}^{2} - 1\right)$</p><p>$\tan^{2}\theta = \sec^{2}\theta\left({e}^{2} - 1\right) ⇒ {e}^{2} - 1 = \sin^{2}\theta$</p><p>${e}^{2} - 1 \in \left[0, 1\right] ∵ \theta \in \left(0, \dfrac{\pi}{2}\right)$</p><p>${e}^{2} \in \left[1, 2\right] ⇒ 1 < e < \sqrt{2}$</p><p>Now$A\left(1, 0\right)$P\left(asec\theta, btan\theta\right) = \left(se{c}^{2}\theta, \tan^{2}\theta\right)$

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

$∵ ∆ APB$ is isosceles-right angled triangle,

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