Question

Find the equations of the tangent and normal to the curve $y=6-{{x}^{2}}$, where normal is parallel to the line $x-4y+3=0$. \[\]

Answer 1

Hint: We find the slope of the normal which is equal to the slope parallel line $x-4y+3=0$ as ${{m}_{n}}$. We find the slope of the tangent line ${{m}_{t}}=\dfrac{-1}{{{m}_{n}}}$. We differentiate the given curve with respect to $x$ and equate the obtained expression for slope of at any point with ${{m}_{t}}$ to get $x-$ coordinate of point of contact. We put $x-$ coordinate in $y=6-{{x}^{2}}$ to get $y-$ coordinate point of contact. We find equation of tangent and normal by slope-point equation of line which passes through $\left( {{x}_{1}},{{y}_{1}} \right)$ with slope $m$ as $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$\[\]

Complete step by step answer:
We know that equation of the slope of the line $ax+by+c=0$ is $m=\dfrac{-a}{b}$.\[\]
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 tangent at that point.
We are given the equation of curve
\[y=6-{{x}^{2}}........\left( 1 \right)\]
Let us differentiate the above curve with respect to $x$ and have the slope of the curve at any point on the curve as
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 6-{{x}^{2}} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=0-2x \\
 & \Rightarrow \dfrac{dy}{dx}=-2x......\left( 2 \right) \\
\end{align}\]
We are further given in the question the equation of a line parallel to normal line
\[x-4y+3=0\]
We compare the equation of normal line with the equation $ax+by+c=0$ and find the slope of the normal since parallel lines have equal slopes as
\[{{m}_{n}}=\dfrac{-a}{b}=\dfrac{-1}{-4}=\dfrac{1}{4}\]
We know that the product of slopes of two perpendicular lines is always $-1$. Since normal line and tangent line at any particular point are always perpendicular, we have the slope of the tangent as
\[{{m}_{t}}=\dfrac{-1}{{{m}_{n}}}=\dfrac{-1}{\dfrac{1}{4}}=-4\]
Now we need to find the point of contact. We equate the slope at any point of the curve from (2) to the slope of tangent to have;
\[\begin{align}
  & -2x=-4 \\
 & \Rightarrow x=2 \\
\end{align}\]
We put in equation (1) to have;
\[y=6-{{x}^{2}}=6-{{\left( 2 \right)}^{2}}=2\]
So the point of contact of the tangent and the curve is $\left( 2,2 \right)$ and the slope is ${{m}_{t}}=-4$. We use slope-point of equation of line and have the equation of tangent as;
\[\begin{align}
  & y-2=-4\left( x-2 \right) \\
 & \Rightarrow y-2=-4x+8 \\
 & \Rightarrow 4x+y-10=0 \\
\end{align}\]
We again use slope-point form of equation of line and have the equation of normal which passes through $\left( 2,2 \right)$ and has slope ${{m}_{n}}=\dfrac{-1}{4}$ as,
\[\begin{align}
  & y-2=\dfrac{1}{4}\left( x-2 \right) \\
 & \Rightarrow 4y-8=x-2 \\
 & \Rightarrow x-4y+6=0 \\
\end{align}\]

Note: We note that the given curve $y=6-{{x}^{2}}$ is an equation of downward parabola which can have only one tangent at on point. The equation of tangent to the downward parabola ${{x}^{2}}=4ay, a < 0 $ at any point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by $x{{x}_{1}}=2a\left( y+{{y}_{1}} \right)$ and with slope of parabola $m$ is given by $y=mx-a{{m}^{2}}$ at the p[point $\left( 2am,a{{m}^{2}} \right)$.