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

Question

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

Aicharya · Accepted Answer

$y=\frac{x}{x^2-3} \Rightarrow \frac{d y}{d x}=\frac{-\left(3+x^2ight)}{\left(x^2-3ight)^2}$ Now slope of $2 x+6 y-11=0$ is $\frac{-1}{3}$ $\begin{aligned} & \frac{\mathrm{A}}{\mathrm{Q}}-\frac{1}{3}=\frac{-\left(3+x^2ight)}{\left(x^2-3ight)^2} \ & \Rightarrow x^4-9 x^2=0 \ & \Rightarrow x^2\left(x^2-9ight)=0 \ & \Rightarrow x=0 	ext { or } x= \pm 3\end{aligned}$ But $x 
eq 0$ so $x= \pm 3 \Rightarrow y= \pm \frac{1}{2}$ Hence, $\alpha= \pm 3, \beta= \pm \frac{1}{2}$ $\Rightarrow|6 \alpha+2 \beta|=\left| \pm\left(6 	imes 3+2 	imes \frac{1}{2}ight)ight|=19$

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)…

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