Question

Find the equations of the tangent and the normal, to the curve $16x^2+y^2=145$ at the point $(x_1,y_1)$, where $x_1=2$ and $y_1>0$.

Answer 1

On differentiating the curve wrt $x$, $32x+2y\frac{dy}{dx}=0$

$⟹{\frac{dy}{dx}}_{(x_1,y_1)}=-\frac{16x_1}{y_1}$

$\therefore$ tangent at $(x_1,y_1)$ would be $\frac{y-y_1}{x-x_1}=-\frac{16x_1}{y_1}$

$⟹y-y_1=-\frac{32}{y_1}(x-2)⟹y{y}_1+32x=y_1^2+64$

$\therefore$ tangent at $(x_1,y_1)$ would be $\frac{y-y_1}{x-x_1}=\frac{y_1}{16x_1}$

$⟹y-y_1=\frac{y_1}{32}(x-2)$

$⟹32y-y_{1}x=32y_1-2y_1$

These are the required solutions.