Question:
If the tangent to the curve $\mathrm{y}=\frac{\mathrm{x}}{\mathrm{x}^{2}-3}, \mathrm{x} \in \mathrm{R}$,
$(x \neq \pm \sqrt{3})$, at a point $(\alpha, \beta) \neq(0,0)$ on it is parallel
to the line $2 x+6 y-11=0$, then :
Correct Option: 1
Solution:
$\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(\alpha, \beta)}=\frac{-\alpha^{2}-3}{\left(\alpha^{2}-3\right)^{2}}$
Given that :
$\frac{-\alpha^{2}-3}{\left(\alpha^{2}-3\right)^{2}}=-\frac{1}{3}$
$\Rightarrow \alpha=0, \pm 3 \quad(\alpha \neq 0)$
$\Rightarrow \beta=\pm \frac{1}{2}, \quad(\beta \neq 0)$
$|6 \alpha+2 \beta|=19$