Let a tangent to the curve 9 x ^ 2 + 16 y ^ 2 = 144 intersect the...

Given:

$9{x}^{2} + 16{y}^{2} = 144$

$⇒ \dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{9} = 1$

Equation of tangent at point $P\left(4cos\theta, 3sin\theta\right) is </p><p>\dfrac{xcos\theta}{4} + \dfrac{ysin\theta}{3} =$1$</p><p>So, A$ is \left(4sec\theta, 0\right)$and point$B$is \left(0, 3 cosec \theta\right)$.

Hence,

$AB = \sqrt{16\sec^{2}\theta + 9{cosec}^{2}\theta}$

$⇒ AB = \sqrt{25 + 16\tan^{2}\theta + 9\cot^{2}\theta}$

$⇒ \sqrt{25 + 16\tan^{2}\theta + 9\cot^{2}\theta} \ge 7$