Question

Find the equation for all lines having slopes $2$ and being tangent to the curve $y+\frac{2}{x-3}=0$.

Answer 1

Consider the given equation,

$y+\frac{2}{x-3}=0$ …..(1)

Differentiate with respect to x , we get

$\frac{dy}{dx}+\frac{d}{dx}\left(\frac{2}{x-3}\right)=\frac{d}{dx}0$

$\frac{dy}{dx}+2(-1)\times \frac{1}{{(x-3)}^2}=0$

$\frac{dy}{dx}=\frac{2}{{(x-3)}^2}$

But given that $\frac{dy}{dx}=2$

$2=\frac{2}{{(x-3)}^2}$

$1=\frac{1}{{(x-3)}^2}$

${(x-3)}^2=1$

$x-3=\pm 1$

$x=4,x=2$

When,$x=4$ put in equation 1^st we get $y=-2$

When, $x=2$,then$y=2$

Now, if $(x,y)=(4,-2)$ $$

$\begin{align}

$y-(-2)=2(x-4)$

$y+2=2x-8$

$2x-y=10$