Application of Derivatives 1 Question 1
####1. 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
(2019 Main, 10 April II)
(a) $|6 \alpha+2 \beta|=19$
(b) $|6 \alpha+2 \beta|=9$
(c) $|2 \alpha+6 \beta|=19$
(d) $|2 \alpha+6 \beta|=11$
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Answer:
Correct Answer: 1. (a)
Solution:
- Equation of given curve is
$ y=\frac{x}{x^{2}-3}, x \in R,(x \neq \pm \sqrt{3}) $
On differentiating Eq. (i) w.r.t. $x$, we get
$ \frac{d y}{d x}=\frac{\left(x^{2}-3\right)-x(2 x)}{\left(x^{2}-3\right)^{2}}=\frac{\left(-x^{2}-3\right)}{\left(x^{2}-3\right)^{2}} $
It is given that tangent at a point $(\alpha, \beta) \neq(0,0)$ on it is parallel to the line
$ 2 x+6 y-11=0 . $
$\therefore$ Slope of this line $=-\frac{2}{6}=\left.\frac{d y}{d x}\right|_{(\alpha, \beta)}$
$\Rightarrow \quad-\frac{\alpha^{2}+3}{\left(\alpha^{2}-3\right)^{2}}=-\frac{1}{3}$
$\Rightarrow \quad 3 \alpha^{2}+9=\alpha^{4}-6 \alpha^{2}+9$
$\Rightarrow \quad \alpha^{4}-9 \alpha^{2}=0$
$\Rightarrow \quad \alpha=0,-3,3$
$\Rightarrow \quad \alpha=3$ or $-3, \quad[\because \alpha \neq 0]$
Now, from Eq. (i),
$ \beta=\frac{\alpha}{\alpha^{2}-3} \Rightarrow \beta=\frac{3}{9-3} \text { or } \frac{-3}{9-3}=\frac{1}{2} \text { or }-\frac{1}{2} $
According to the options, $|6 \alpha+2 \beta|=19$
at $(\alpha, \beta)= \pm 3, \pm \frac{1}{2}$
