Search any question & find its solution
Question:
Answered & Verified by Expert
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 $2 x+6 y-11=0$, then
Options:
Solution:
1752 Upvotes
Verified Answer
The correct answer is:
$|6 \alpha+2 \beta|=19$
$y=\frac{x}{x^2-3} \Rightarrow \frac{d y}{d x}=\frac{-\left(3+x^2\right)}{\left(x^2-3\right)^2}$
Now slope of $2 x+6 y-11=0$ is $\frac{-1}{3}$
$\begin{aligned} & \frac{\mathrm{A}}{\mathrm{Q}}-\frac{1}{3}=\frac{-\left(3+x^2\right)}{\left(x^2-3\right)^2} \\ & \Rightarrow x^4-9 x^2=0 \\ & \Rightarrow x^2\left(x^2-9\right)=0 \\ & \Rightarrow x=0 \text { or } x= \pm 3\end{aligned}$
But $x \neq 0$ so $x= \pm 3 \Rightarrow y= \pm \frac{1}{2}$
Hence, $\alpha= \pm 3, \beta= \pm \frac{1}{2}$
$\Rightarrow|6 \alpha+2 \beta|=\left| \pm\left(6 \times 3+2 \times \frac{1}{2}\right)\right|=19$
Now slope of $2 x+6 y-11=0$ is $\frac{-1}{3}$
$\begin{aligned} & \frac{\mathrm{A}}{\mathrm{Q}}-\frac{1}{3}=\frac{-\left(3+x^2\right)}{\left(x^2-3\right)^2} \\ & \Rightarrow x^4-9 x^2=0 \\ & \Rightarrow x^2\left(x^2-9\right)=0 \\ & \Rightarrow x=0 \text { or } x= \pm 3\end{aligned}$
But $x \neq 0$ so $x= \pm 3 \Rightarrow y= \pm \frac{1}{2}$
Hence, $\alpha= \pm 3, \beta= \pm \frac{1}{2}$
$\Rightarrow|6 \alpha+2 \beta|=\left| \pm\left(6 \times 3+2 \times \frac{1}{2}\right)\right|=19$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.