Question

If the tangent to the curve $y=\frac{x}{x^2-3},x\in R,(x\ne \pm \sqrt{3},)$ at a point $(\alpha ,\beta )\ne (0,0)$ is parallel to the line $2x+6y-11=0,$ then $|6\alpha +2\beta |=$

Answer 1

$y=\frac{x}{x^2-3}...\left(1\right)$
Differentiating w.r.t $x$, we get
$\frac{dy}{dx}=\frac{-x^2-3}{(x^2-3{)}^2}$
$\frac{dy}{dx}{|}_{(\alpha ,\beta )}=\frac{-{\alpha }^2-3}{({\alpha }^2-3{)}^2}$
The slope of line $=\frac{dy}{dx}=-\frac{1}{3}$
$\Rightarrow \frac{-{\alpha }^2-3}{({\alpha }^2-3{)}^2}=-\frac{1}{3}$
$\Rightarrow \alpha =0,\pm 3(\alpha \ne 0)$
Now, from equation $\left(1\right)$
$\beta =\frac{\alpha }{{\alpha }^2-3}=\pm \frac{1}{2}$
$(\alpha ,\beta )=(3,\frac{1}{2}),(-3,-\frac{1}{2})$
$\Rightarrow |6\alpha +2\beta |=19$