Question

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

• A. $|6 \alpha + 2 \beta | = 19$
• B. $|2 \alpha + 6 \beta | = 11 $
• C. $|6 \alpha + 2 \beta | = 9$
• D. $|2 \alpha + 6 \beta | = 19$

Answer 1

Answer: (A)

$\frac{dy}{dx} \Bigg|_{\left(\alpha , \beta\right)} = \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 \pm3 \left(\alpha \ne0 \right) $
$ \Rightarrow \beta=\pm \frac{1}{2} . \left(\beta\ne0\right) $
$ \left|6\alpha+2\beta\right|=19 $