The tangent to the ellipse x^ 2 25 + y^ 2 16 = 1 at point P lying...

Let point $P$ be \left(5cos \theta, 4sin \theta\right), \theta \in \left(0, \dfrac{\pi}{2}\right)$<br />So, the equation of tangent is$\dfrac{xcos \theta}{5} + \dfrac{ysin \theta}{4} = 1$<br />Points$Q$and$R$are \left(5sec \theta, 0\right)$ and $\left(0, 4cosec \theta\right)$ respectively.
Hence,~{\left(QR\right)}^{2} = 25\sec^{2} \theta + 16{cosec}^{2} \theta = 25 \left(1 + \tan^{2} \theta\right) +16 \left(1 + \cot^{2} \theta\right)<br />= 41 + 25\tan^{2} $\theta +$16\cot^{2} $\theta =$41 + ${ \left(5tan \theta -4cot \theta\right)}^{2} + 40$
{\left(QR\right)}_\min^{2} = 81 when $\tan^{2} \theta = \dfrac{4}{5}⇒ tan \theta = \dfrac{2}{\sqrt{5}}$
Therefore,~$sin \theta = \dfrac{2}{3},cos \theta = \dfrac{\sqrt{5}}{3}$
Hence, equation of tangent is $2x + \sqrt{5}y = 6\sqrt{5}$