Let P be a point in the first quadrant lying on the ellipse $9{\mathrm{x}}^2+16{\mathrm{y}}^2=144,$ such that the tangent at P to the ellipse is inclined at an angle ${135}^0$ to the positive direction of x-axis. Then the coordinates of P are

Given ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}=1(\mathrm{a}>\mathrm{b})$

where ${\mathrm{a}}^2=16, {\mathrm{b}}^2=9$

Let $\mathrm{P}=(4\mathrm{cos\theta }, 3\mathrm{sin\theta })$ be any point on ellipse.

Equation of tangent at $\mathrm{P}\left(\mathrm{\theta }\right)$ is

$\frac{\mathrm{x}}{4}\mathrm{cos\theta }+\frac{\mathrm{y}}{3}\mathrm{sin\theta }=1$

slope of tangent (m) $=\frac{{\displaystyle \frac{-\mathrm{cos\theta }}{4}}}{{\displaystyle \frac{\mathrm{sin\theta }}{3}}}$

$\Rightarrow \mathrm{tan\theta }=\frac{-\mathrm{cos\theta }}{4}\times \frac{3}{\mathrm{sin\theta }}$

$\Rightarrow \tan {135}^0=\frac{-3}{4} \mathrm{cot\theta }$    ($\therefore$ inclination of tangent $={135}^0$ )

$$\begin{gathered} \Rightarrow -1=\frac{-3}{4} \mathrm{cot\theta } \\ \Rightarrow \mathrm{tan\theta }=\frac{3}{4} \end{gathered}$$                                          

Now  $\mathrm{sin\theta }=\frac{3}{5}, \mathrm{cos\theta }=\frac{4}{5}$

$$\begin{gathered} \therefore \mathrm{P}=\left(4\left(\frac{4}{5}\right), 3\left(\frac{3}{5}\right)\right) \\ =\left(\frac{16}{5}, \frac{9}{5}\right) \end{gathered}$$