Question

If the tangent to the curve $y=\frac{x}{x^2-3},x\in R,(x\ne \pm \sqrt{3},)$ at a point $(\alpha ,\beta )\ne (0,0)$ 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

The correct option is A $|6\alpha +2\beta |=19$

$y=\frac{x}{x^2-3}...\left(1\right)$
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 is equal to $\frac{dy}{dx}$ which is $-\frac{1}{3}$
$\Rightarrow \frac{-{\alpha }^2-3}{({\alpha }^2-3{)}^2}=-\frac{1}{3}$
$\Rightarrow \alpha =0,\pm 3(\alpha \ne 0)$

Now from equation $\left(1\right)$
$\beta =\frac{\alpha }{{\alpha }^2-3}=\pm \frac{1}{2}$

$(\alpha ,\beta )=(3,\frac{1}{2}),(-3,-\frac{1}{2})$
$\Rightarrow |6\alpha +2\beta |=19$