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

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 $
$\Rightarrow a=\pm3, \beta=\pm \frac{1}{2}$
These values of $\alpha$ and $\beta$ satisfies $\left|6 \alpha+2\beta\right|=19$