Question

Find the equation of normal to the curve $3{{x}^{2}}-{{y}^{2}}=8$ which is parallel to the line $x+3y=4$.\[\]

Answer 1

Hint: Differentiate the given equation of curve $3{{x}^{2}}-{{y}^{2}}=8$ with respect to $x$ to find the slope of tangent and then normal . Find the slope of the normal $\left( m \right)$ from the given parallel line and equate it with the obtained expression to find $y$ in terms of $x$. Put that in the equation of the curve to find possible points $\left( {{x}_{1}},{{y}_{1}} \right)$ say of intersection of the normal and curve. Check which are admissible points and use slope-point equation $\left( y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \right)$. \[\]

Complete step-by-step answer:
We know from differential calculus that the slope of any curve at any point is given by the differentiation with respect to the independent variable. We also know that the slope of the curve at any point is the slope of the tangent at that point. \[\]
The equation of the given curve is
\[3{{x}^{2}}-{{y}^{2}}=8...(1)\]
Let us differentiate the above equation with respect to $x$, the independent variable. n
\[\begin{align}
  & \dfrac{d}{dx}\left( 3{{x}^{2}}-{{y}^{2}} \right)=\dfrac{d}{dx}8. \\
 & \Rightarrow 6x-2y\dfrac{dy}{dx}=0 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{3x}{y} \\
\end{align}\]
We also know that the product of slopes of the normal and the tangent to a curve as just as any pair of orthogonal lines is $-1.$ So the slope of the normal at any point is $m=\dfrac{-1}{\dfrac{dy}{dx}}=\dfrac{-1}{\dfrac{3x}{y}}=\dfrac{-y}{3x}$.\[\]
 We know that the slopes of two parallel lines are equal. It is given that the line $x+3y=4$ is parallel to the normal to the curve. We find the slope of the given line by converting it to the form slope-intercept form ,
\[\begin{align}
  & x+3y=4 \\
 & \Rightarrow y=\dfrac{1}{3}x-4 \\
\end{align}\]
So the slope of the line and the normal is $\dfrac{1}{3}$ which implies $\dfrac{-y}{3x}=\dfrac{-1}{3}\Rightarrow y=x$. We put it in equation (1) and get,
\[\begin{align}
  & 3{{x}^{2}}-{{x}^{2}}=8 \\
 & \Rightarrow {{x}^{2}}=4 \\
 & \Rightarrow x=2,-2 \\
\end{align}\]
We put the values of $x$ in equation (1) and get two points on the normal.
\[\begin{align}
  & {{y}^{2}}=\dfrac{3{{x}^{2}}-4}{3} \\
 & \Rightarrow {{y}^{2}}=\dfrac{3{{\left( \pm 2 \right)}^{2}}-4}{3}=4 \\
 & \Rightarrow y=2,-2 \\
\end{align}\]

So possible the points at which the normal cuts the curve is $\left( \pm 2,\pm 2 \right)$. We know the slope of normal is $m=\dfrac{-y}{3x}=\dfrac{-1}{3}$ which only satisfies on the points$\left( 2,2 \right),\left( -2,-2 \right)$. We use the slope point equation of normal using the points $\left( 2,2 \right)$ and $\left( -2,-2 \right)$
\[\begin{align}
  & y-2=\dfrac{-1}{3}\left( x-2 \right)\Rightarrow 3y+x=8 \\
 & y-\left( -2 \right)=\dfrac{-1}{3}\left( x-\left( -2 \right) \right)\Rightarrow 3y+x=-8 \\
\end{align}\]
The above linear equations are the required equations of normal. \[\]

Note: We need to be careful to select admissible points for the equation of normal. We note that the given equation is an equation of hyperbola of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. We can directly find the pair of tangents $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$ where $m$ is the slope. We can also get equation of normal at a point $\left( {{x}_{1}},{{y}_{1}} \right)$ as $\dfrac{{{a}^{2}}x}{{{x}_{1}}}-\dfrac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}-{{b}^{2}}$.