If the tangent to the curve

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 :

  1. $|6 \alpha+2 \beta|=19$

  2. $|2 \alpha+6 \beta|=11$

  3. $|6 \alpha+2 \beta|=9$

  4. $|2 \alpha+6 \beta|=19$


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 :

$\frac{-\alpha^{2}-3}{\left(\alpha^{2}-3\right)^{2}}=-\frac{1}{3}$

$\Rightarrow \alpha=0, \pm 3 \quad(\alpha \neq 0)$

$\Rightarrow \beta=\pm \frac{1}{2}, \quad(\beta \neq 0)$

$|6 \alpha+2 \beta|=19$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now