Question

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

Answer 1

Answer: 19

Given curve is, $y=\frac{x}{x^2-3}$
$\Rightarrow \frac{dy}{dx}=\frac{\left(x^2-3\right)-x\left(2x\right)}{{\left(x^2-3\right)}^2}=\frac{-x^2-3}{{\left(x^2-3\right)}^2}$
$\frac{dy}{dx}{|}_{\left(\alpha ,\beta \right)}=\frac{{\alpha }^2-3}{{\left({\alpha }^2-3\right)}^2}=-\frac{2}{6}=-\frac{1}{3}$
$3\left({\alpha }^2+3\right)={\left({\alpha }^2-3\right)}^2$
$\Rightarrow {\alpha }^2=9$
And, $\beta =\frac{x}{a^2-3}$
$\Rightarrow {\alpha }^2-3=\frac{\alpha }{\beta }$
$\Rightarrow \frac{\alpha }{\beta }=6$
$\Rightarrow a=\pm 3,\beta =\pm \frac{1}{2}$
These values of $\alpha$ and $\beta$ satisfies $\left|6\alpha +2\beta \right|=19$