Question

If the tangent to the curve y=$\dfrac{x}{{{x}^{2}}-3},\text{ }x\in \mathbb{R},\text{ }\left( x\ne \pm \sqrt{3} \right),\text{ at a point }\left( \alpha ,\beta \right)\ne \left( 0,0 \right)$
$\text{on it is parallel to the line 2x+6y-11=0, then}$
\[\begin{align}
  & \text{A) }\!\!|\!\!\text{ 6}\alpha +2\beta \text{ }\!\!|\!\!\text{ =19} \\
 & \text{B) }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =11} \\
 & \text{C) }\!\!|\!\!\text{ 6}\alpha +2\beta \text{ }\!\!|\!\!\text{ =9} \\
 & \text{D) }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =19} \\
\end{align}\]

Answer 1

Hint: In this, we will compute the value of alpha and beta by using a given curve and line. Firstly we differentiate the given curve and find the slope of the tangent. Also the tangent is parallel to line 2x + 6y – 11=0. By comparing the value slope of tangent and parallel line get the value of alpha and then substituting the value of the alpha in the given curve we get the value of beta. Then we will check each option to see which option satisfies by alpha and beta.

Complete step by step answer:
The given curve is,
$\text{y=}\dfrac{x}{{{x}^{2}}-3},\text{ }x\in \mathbb{R}$
By differentiating y with respect to x, we get
$\dfrac{dy}{dx}\text{=}\dfrac{d}{dx}\left( \dfrac{x}{{{x}^{2}}-3} \right)$
Since, the y is of the form $\dfrac{u}{v}$ where u and v is a function of x. Then derivative of $\dfrac{u}{v}$ is given as
follows
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}.....(2)$
By using formula in equation (2), we get
$\dfrac{dy}{dx}\text{=}\dfrac{d}{dx}\left( \dfrac{x}{{{x}^{2}}-3} \right)=\dfrac{\left( {{x}^{2}}-3 \right)\dfrac{dx}{dx}-x\dfrac{d\left( {{x}^{2}}-3 \right)}{dx}}{{{\left( {{x}^{2}}-3 \right)}^{2}}}$
$\dfrac{dy}{dx}\text{=}\dfrac{\left( {{x}^{2}}-3 \right)(1)-x\left( 2x \right)}{{{\left( {{x}^{2}}-3 \right)}^{2}}}$
$\dfrac{dy}{dx}\text{=}\dfrac{{{x}^{2}}-3-2{{x}^{2}}}{{{\left( {{x}^{2}}-3 \right)}^{2}}}$
$\dfrac{dy}{dx}\text{=}\dfrac{-3-{{x}^{2}}}{{{\left( {{x}^{2}}-3 \right)}^{2}}}$
$\dfrac{dy}{dx}\text{=}\dfrac{-\left( {{x}^{2}}+3 \right)}{{{\left( {{x}^{2}}-3 \right)}^{2}}}$
${{\left. \dfrac{dy}{dx} \right|}_{\left( \alpha ,\beta \right)}}\text{=}\dfrac{-\left( {{\alpha }^{2}}+3 \right)}{{{\left( {{\alpha }^{2}}-3 \right)}^{2}}}$
Which is the slope of tangent to the given curve at point $\left( \alpha ,\beta \right)$
The given equation of line is
2x + 6y – 11=0
 $y=-\dfrac{2}{6}x+\dfrac{11}{6}$
Hence slope of line = $-\dfrac{2}{6}=-\dfrac{1}{3}$
Since the tangent is parallel to line 2x + 6y – 11=0
Therefore, slope of tangent = slope of line
$\dfrac{-\left( {{\alpha }^{2}}+3 \right)}{{{\left( {{\alpha }^{2}}-3 \right)}^{2}}}=-\dfrac{1}{3}$
$-3\left( {{\alpha }^{2}}+3 \right)=-{{\left( {{\alpha }^{2}}-3 \right)}^{2}}$
By cancelling minus on both sides, we get
$3\left( {{\alpha }^{2}}+3 \right)={{\left( {{\alpha }^{2}}-3 \right)}^{2}}$
$3{{\alpha }^{2}}+9={{\alpha }^{4}}-6{{\alpha }^{2}}+9$
By subtracting -9 on both sides, we get
$3{{\alpha }^{2}}={{\alpha }^{4}}-6{{\alpha }^{2}}$
${{\alpha }^{4}}-9{{\alpha }^{2}}=0$
${{\alpha }^{2}}\left( {{\alpha }^{2}}-9 \right)=0$
${{\alpha }^{2}}=0\text{ or }{{\alpha }^{2}}-9=0$
${{\alpha }^{2}}=0\text{ or }{{\alpha }^{2}}=9$
$\alpha =0\text{ or }\alpha =\pm 3$
But $\alpha \ne 0$
$\Rightarrow \alpha =\pm 3$
Since, $\left( \alpha ,\beta \right)$ lies on $\text{y=}\dfrac{x}{{{x}^{2}}-3},\text{ }x\in \mathbb{R}$
$\beta \text{=}\dfrac{\alpha }{{{\alpha }^{2}}-3}$
When $\alpha =3\Rightarrow \beta \text{=}\dfrac{3}{{{3}^{2}}-3}=\dfrac{3}{9-3}=\dfrac{3}{6}=\dfrac{1}{2}$
$\Rightarrow \beta \text{=}\dfrac{1}{2}$
When $\alpha =-3\Rightarrow \beta \text{=}\dfrac{-3}{{{\left( -3 \right)}^{2}}-3}=\dfrac{-3}{9-3}=\dfrac{-3}{6}=-\dfrac{1}{2}$
$\Rightarrow \beta \text{=}-\dfrac{1}{2}$
Now we will check that $\left( \alpha ,\beta \right)=\left( 3,\dfrac{1}{2} \right)=\left( -3,-\dfrac{1}{2} \right)$ which option
Option A, \[\text{ }\!\!|\!\!\text{ 6}\alpha +2\beta \text{ }\!\!|\!\!\text{ =19}\]
For $\left( \alpha ,\beta \right)=\left( 3,\dfrac{1}{2} \right)$
\[\text{LHS= }\!\!|\!\!\text{ 6}\alpha +2\beta \text{ }\!\!|\!\!\text{ =}\left| \text{6(3)+2}\left( \dfrac{1}{2} \right) \right|\text{= }\!\!|\!\!\text{ 18+1 }\!\!|\!\!\text{ =19=RHS}\]
For $\left( \alpha ,\beta \right)=\left( -3,-\dfrac{1}{2} \right)$
\[\text{LHS= }\!\!|\!\!\text{ 6}\alpha +2\beta \text{ }\!\!|\!\!\text{ =}\left| \text{6}\left( -3 \right)\text{+2}\left( -\dfrac{1}{2} \right) \right|\text{= }\!\!|\!\!\text{ -18-1 }\!\!|\!\!\text{ =19=RHS}\]
Hence both valve of $\left( \alpha ,\beta \right)$ satisfies option A
By looking in option A satisfied by both the value of $\left( \alpha ,\beta \right)$ implies both the value of $\left( \alpha ,\beta \right)$
Does not satisfies option C
 Option B, \[\text{ }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =11}\]
For $\left( \alpha ,\beta \right)=\left( 3,\dfrac{1}{2} \right)$
\[\text{LHS= }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =}\left| \text{2(3)+6}\left( \dfrac{1}{2} \right) \right|\text{= }\!\!|\!\!\text{ 6+3 }\!\!|\!\!\text{ =9}\ne \text{11=RHS}\]
For $\left( \alpha ,\beta \right)=\left( -3,-\dfrac{1}{2} \right)$
\[\text{LHS= }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =}\left| 2\left( -3 \right)\text{+6}\left( -\dfrac{1}{2} \right) \right|\text{= }\!\!|\!\!\text{ -6-3 }\!\!|\!\!\text{ =9}\ne \text{11=RHS}\]
Hence both valve of $\left( \alpha ,\beta \right)$ does not satisfies option B
Option D, \[\text{ }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =19}\]
For $\left( \alpha ,\beta \right)=\left( 3,\dfrac{1}{2} \right)$
\[\text{LHS= }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =}\left| \text{2(3)+6}\left( \dfrac{1}{2} \right) \right|\text{= }\!\!|\!\!\text{ 6+3 }\!\!|\!\!\text{ =9}\ne \text{19=RHS}\]
For $\left( \alpha ,\beta \right)=\left( -3,-\dfrac{1}{2} \right)$
\[\text{LHS= }\!\!|\!\!\text{ 2}\alpha +6\beta \text{ }\!\!|\!\!\text{ =}\left| 2\left( -3 \right)\text{+6}\left( -\dfrac{1}{2} \right) \right|\text{= }\!\!|\!\!\text{ -6-3 }\!\!|\!\!\text{ =9}\ne \text{19=RHS}\]
Hence both valve of $\left( \alpha ,\beta \right)$ does not satisfies option B

So, the correct answer is “Option A”.

Note: In this problem, one should know that if two lines are parallel then their slopes are equal. Always remember that $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$, where u and v are functions of x. if two lines in a plane are perpendicular, then product of their slopes is equal to -1. Try not to make any calculation mistakes.