Question

Find the equation of tangent to the curve $4{{x}^{2}}+9{{y}^{2}}=36$ at the point $\left( 3\cos \theta ,2\sin \theta \right)$ where $\theta $ is the polar angle.

Answer 1

Hint: We differentiate the given equation of curve $4{{x}^{2}}+9{{y}^{2}}=36$ with respect to $x$ and find the slope of the tangent at point $\dfrac{dy}{dx}$. We put the given point $\left( 3\cos \theta ,2\sin \theta \right)$ in the expression of $\dfrac{dy}{dx}$ and find the slope at that point as $m$. We use the slope-point form of the equation of line $y=mx+c$ to obtain the required equation of tangent.

Complete step-by-step solution
We are given the question of the curve
\[4{{x}^{2}}+9{{y}^{2}}=36\]
We know that we can convert any Cartesian coordinate $\left( x,y \right)$ into polar coordinates $\left( a\cos \theta ,b\sin \theta \right)$ where $\sqrt{{{a}^{2}}+{{b}^{2}}}$ is the distance from the origin and $\theta $ is the angle the line joining the point to the origin makes with $x-$ axis called polar angle . \[\]
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. So let us differentiate the given curve with respect to $x$ and find the expression for slope. We have,
\[\begin{align}
  & \dfrac{d}{dx}\left( 4{{x}^{2}}+9{{y}^{2}} \right)=\dfrac{d}{dx}\left( 36 \right) \\
 & \Rightarrow 8x+18y\dfrac{dy}{dx}=0 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-8x}{18y}= \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-4x}{9y} \\
\end{align}\]
We are asked to find the equation of tangent at the point $\left( 3\cos \theta ,2\sin \theta \right)$. So let us put the point $\left( 3\cos \theta ,2\sin \theta \right)$ in the expression for slope and have,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-4x}{9y}=\dfrac{-4\times 3\cos \theta }{9\times 2\sin \theta }=\dfrac{-2\cos \theta }{3\sin \theta }\]
We know that the equation of line with slope $m$ and a point on the line $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by the slope-point from of equation of line
\[y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)+c\]
We are given a point on the tangent line $\left( 3\cos \theta ,2\sin \theta \right)$ in the question and we have obtained the slope $m=\dfrac{dy}{dx}=\dfrac{-2\cos \theta }{3\sin \theta }$. So the equation of tangent line in slope point from is
\[\begin{align}
  & y-2\sin \theta =\dfrac{-2\cos \theta }{3\sin \theta }\left( x-3\cos \theta \right) \\
 & \Rightarrow 3y\sin \theta -6{{\sin }^{2}}\theta =-2x\cos \theta +6{{\sin }^{2}}\theta \\
 & \Rightarrow 2x\cos \theta +3y\sin \theta =6{{\sin }^{2}}\theta+6{{\cos }^{2}}\theta \\
 & \Rightarrow 2x\cos \theta +3y\sin \theta =6\left( {{\sin }^{2}}\theta+{{\cos }^{2}}\theta \right) \\
\end{align}\]
We use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the above step and proceed to have equation of tangent as,
\[\Rightarrow 2x\cos \theta +3y\sin \theta =6\]

Note: We note that when $\theta =0$ the line is parallel to the $y-$axis and when $\theta =\dfrac{\pi }{2}$ the line is parallel to the $x-$axis. The given equation is an equation of ellipse which has standard form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ whose equation at $\left( a\cos \theta ,b\sin \theta \right)$ is given by $\dfrac{x}{a}\cos \theta +\dfrac{y}{a}\sin \theta =1$. The parametric form of any point on the ellipse is given by $\left( a\cos \theta ,b\sin \theta \right)$.