Let P(asecθ, btanθ) and Q(asecϕ, btanϕ) where θ + ϕ = (π)/(2), be...

We have,

$\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1 . . . \left(i\right)$

On differentiating w.r.t., $x,$ we get

$\dfrac{2x}{{a}^{2}} - \dfrac{2y}{{b}^{2}} \times \dfrac{dy}{dx} = 0$

$⇒ \dfrac{dy}{dx} = \dfrac{{b}^{2}}{{a}^{2}}\left(\dfrac{x}{y}\right)$

$⇒ - \dfrac{dx}{dy} = - \dfrac{{a}^{2}}{{b}^{2}}\left(\dfrac{y}{x}\right)$

Slope for normal at the point $P\left(asec\theta, btan\theta\right)$ is

${\left(- \dfrac{dx}{dy}\right)}_{P} = - \dfrac{{a}^{2}btan\theta}{{b}^{2}asec\theta} = - \dfrac{a}{b}sin\theta$

$∴$ Equation of normal at \left(asec\theta, btan\theta\right)$is</p><p>$y - btan\theta = - \dfrac{a}{b}sin\theta\left(x - asec\theta\right)$</p><p>$⇒ \left(asin\theta\right)x + by = \left({a}^{2} + {b}^{2}\right)tan\theta$</p><p>$⇒ ax + b cosec \thetay = \left({a}^{2} + {b}^{2}\right)sec\theta . . . \left(ii\right)$</p><p>Similarly, equation of normal to$\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$at</p><p>$( asecϕ, btanϕ )$is</p><p>$ax + by cosec ϕ = \left({a}^{2} + {b}^{2}\right)secϕ . . . \left(ii\right)$</p><p>On subtracting \left(ii\right)$ from \left(i\right)$, we get</p><p>$b ( cosec \theta - cosec ϕ ) y = \left({a}^{2} + {b}^{2}\right) ( sec\theta - secϕ )$</p><p>$⇒ y = \dfrac{{a}^{2} + {b}^{2}}{b} \cdot \left(\dfrac{sec\theta - secϕ}{cosec \theta - cosec ϕ}\right)$</p><p>But$\dfrac{sec\theta - secϕ}{cosec \theta - cosec ϕ} = \dfrac{sec\theta - \sec \left(\dfrac{\pi}{2} - \theta\right)}{cosec \theta - cosec \left(\dfrac{\pi}{2} - \theta\right)}

\left[∵ ϕ + \theta = \dfrac{\pi}{2}\right]$</p><p>$= \dfrac{sec\theta - cosec \theta}{cosec\theta - sec\theta} = - 1$</p><p>Thus,</p><p>$y = - \left(\dfrac{{a}^{2} + {b}^{2}}{b}\right),$i.e.,$k = 2.