Question

Find the equations of the tangent and the normal, to the curve $16x^2+9y^2=145$ at the point $(x_1,y_1)$, where $x_1=2$ and $y_1>0$.

Answer 1

Given:
The curve, $16x^2+9y^2=145$ ..... $\left(i\right)$

$(x_1,y_1)$ is a point on $\left(i\right)$

$\therefore 16x_1^2+9y_1^2=145$

If $x_1=2$, then

$16(2{)}^2+9y_1^2=145$

$9y_1^2=145-64$

$y_1^2=\frac{81}{9}=9$

$y_1=3[\because y_1>0]$

$\therefore$ The point $(2,3)$ lies on $\left(i\right)$.

Differentiating eqn. $\left(i\right)$ w.r.t. $x$, we have

$32x+18y\cdot \frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{32x}{18y}$

$\frac{dy}{dx}{]}_{at(2,3)}=-\frac{32\times 2}{18\times 3}=-\frac{32}{27}$

Thus, Slope of the tangent $=-\frac{32}{27}$

Slope of normal = $-\frac{1}{\text{Slope of tangent}}$

So, Slope of the normal $=\frac{27}{32}$.

Now, equation of the tangent at $(2,3)$ is :
$y-3=-\frac{32}{27}(x-2)$

$27y-81=-32x+64$

$\Rightarrow 32x+27y-145=0$

Equation of the normal at $(2,3)$ is :

$y-3=\frac{27}{32}(x-2)$

$32y-96=27x-54$

$\Rightarrow 27x-32y+42=0$