Find the equations of the tangent and the normal to `16 x^2+9y^2=144` at `(x_1,\ y_1)` where `x_1=2` and `y_1>0` .

Answer

`8 x+3 sqrt{5} y-36=0`

Explanation

AS Given;
$$
\begin{aligned}
&16 \mathrm{x}^{2}+9 \mathrm{y}^{2}=144 \rightarrow(\mathrm{i}) \\
&\text { since, }\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { lie on eq(i) } \\
&\Rightarrow 16 \mathrm{x}_{1}^{2}+9 \mathrm{y}_{1}^{2}=144 \\
&\mathrm{x}_{1}=2 \\
&\Rightarrow 16 \times(2)^{2}+9 \mathrm{y}_{1}^{2}=144 \\
...