Question

Find the equation of tangent to the curve $16x^2+9y^2=144$ at the point $(x_1,y_2)$ where $x_1=2$ and $y_1>0$.

Answer 1

$\begin{array}{c}16x^2+9y^2=144\to \left(i\right)\\ \sin ce,\left(x_1,y_1\right)lieoneq\left(i\right)\\ \Rightarrow 16x_1^2+9y_1^2=144\\ x_1=2\\ \Rightarrow 16\times {\left(2\right)}^2+9y_1^2=144\\ \Rightarrow y_1\frac{4\sqrt{5}}{3}\left[y_1>0\right]\end{array}$

so, the coordinates of given point is $\left[2,\frac{4\sqrt{5}}{3}\right]$

$\begin{array}{c}now,\\ 16x^2+9y^2=144\\ \Rightarrow 32x+18y\frac{dy}{dx}=0\\ differentiatingw.r.to.x\\ \Rightarrow \frac{dy}{dx}=\frac{-16}{9y}\\ \Rightarrow \frac{dy}{dx}\left(2,\frac{4\sqrt{5}}{3}\right)=\frac{-8}{3\sqrt{5}}=m\end{array}$

equation of tangent with slope m at

$\begin{array}{c}2,\left(\frac{4\sqrt{5}}{3}\right)\\ \Rightarrow y-\frac{-4\sqrt{5}}{3}=m\left(x-2\right)\\ \Rightarrow 8x+3\sqrt{5}y-36=0\end{array}$