Parametric Equation of an Ellipse

Standard Equation of Ellipse

The standard form of the equation of an ellipse with center $(0,0)$ and major axis on the $x$-axis is $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathbf{y}^2}{\mathbf{b}^2}=1 \quad$ where, $\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)$
1. $a>b$
2. the length of the major axis is $2 a$
3. the length of the minor axis is $2 b$
4. the coordinates of the vertices are $( \pm a, 0)$

What is the Parametric equation of Ellipse?

The equations $x=a \cos \theta, y=b \sin \theta$ are called the parametric equation of the ellipse.

The parametric equation of the ellipse is given by $x=a \cos \theta, y=b \sin \theta$ and the parametric coordinates of the points lying on it is ( $a \cos \theta, b \sin \theta)$.

Derivation of Parametric Equation of Ellipse

Let $P(x, y)$ be a point on the ellipse
Draw PN perpendicular to the major axis and PN to meet the auxiliary circle at Q.
Let $\angle \mathrm{ACQ}$ be $\theta$ (This angle is also known as Eccentric Angle). Hence, the parametric equation of circle at point Q (a $\cos \theta$, a $\sin \theta$ ).
Thus, P has x -coordinate as a $\cos \theta$
As P lies on the ellipse

$
\frac{\mathrm{a}^2 \cos ^2 \theta}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \Rightarrow \mathrm{y}= \pm \mathrm{b} \sin \theta
$
Hence, Point P is $(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$

What is Auxillary Circle?

The circle described on the major axis as the diameter is called the auxiliary circle.
Equation of Ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Then, equation of auxiliary circle is $x^2+y^2=a^2$ (As $A A^{\prime}$ is Diameter)

Summary

Parametric equations provide a versatile framework for describing complex curves and paths by expressing coordinates x and y as functions of a parameter t. It enhances the understanding and visualization of mathematical relationships. It has practical applications across various disciplines, from physics and engineering to computer graphics and beyond.

Recommended Video Based on the Parametric Equation of Ellipse

Solved Examples Based on the Parametric equation of Ellipse

Example 1: Let $P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R$, and S be four points on the ellipse $9 x^2+4 y^2=36$. Let PQ and RS be mutually perpendicular and pass through the origin. If $\frac{1}{\mathrm{(PQ})^2}+\frac{1}{(\mathrm{RS})^2}=\frac{\mathrm{p}}{\mathrm{q}}$, where p and q are coprime, then $\mathrm{p}+$ $q$ is equal to $\quad$ JJEE MAINS 2023]
Solution: Let $R(2 \cos \theta, 3 \sin \theta)$
As $\mathrm{OP} \perp \mathrm{OR}$
so $\frac{3 \sin \theta}{2 \cos \theta} \times \frac{\frac{6}{\sqrt{7}}}{\frac{2 \sqrt{3}}{\sqrt{7}}}=-1$

$
\begin{aligned}
& \Rightarrow \tan \theta=\frac{-2}{3 \sqrt{3}} \\
& \Rightarrow \mathrm{R}\left(\frac{-6 \sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}}\right) \text { or } R\left(\frac{6 \sqrt{3}}{\sqrt{31}}, \frac{-6}{\sqrt{31}}\right)
\end{aligned}
$

$\mathrm{Now}=\frac{1}{(\mathrm{PQ})^2}+\frac{1}{(\mathrm{RS})^2}=\frac{1}{4}\left(\frac{1}{(\mathrm{OP})^2}+\frac{1}{(\mathrm{OR})^2}\right)$

$
\begin{aligned}
& =\frac{1}{4}\left(\frac{1}{48}+\frac{1}{144}\right)=\frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right) \\
& =\frac{13}{144} \\
& \Rightarrow p+q=157
\end{aligned}
$
Hence, the answer is 157

Example 2: If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse $x^2+4 y^2=36$ is r, then $12 r^2$ is equal to:
[JEE MAINS 2023]
Solution: C $(2,0)$
Ellipse $x^2+4 y^2=36$

$
\frac{x^2}{36}+\frac{y^2}{9}=1
$
Equation of Normal at $P(6 \cos \theta, 3 \sin \theta)$ is $(6 \sec \theta) x-(3 \operatorname{cosec} \theta) y=27$
It passes through $(2,0)$

$
\Rightarrow \sec \theta=\frac{27}{12}=\frac{9}{4}
$

$\cos \theta \frac{4}{9}, \sin \theta=\frac{\sqrt{65}}{9}$

$
\begin{aligned}
& \mathrm{P}\left(\frac{8}{3}, \frac{\sqrt{65}}{3}\right) \\
& \frac{\gamma}{\mathrm{P}\left(\frac{8}{3}, \frac{\sqrt{65}}{3}\right) \mathrm{c}(2,0)} \\
& \gamma=\sqrt{\left(\frac{8}{3}-2\right)^2+\left(\frac{\sqrt{65}}{3}\right)^2}=\frac{\sqrt{69}}{3}
\end{aligned}
$
Value of $12 \gamma^2=\left(\frac{\sqrt{69}}{3}\right)^2 \times 12$

$
\Rightarrow \frac{12 \times 69}{9}=92
$
Hence, the answer is 92

Example 3: The locus of the midpoint of the line segment joining the point $(4,3)$ and the points on the ellipse $x^2+2 y^2=4$ is an ellipse with eccentricity.
[JEE MAINS 2022]
Solution

$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{2}=1 \\
& p:(2 \cos \theta, \sqrt{2} \sin \theta) \\
& \therefore \quad \hbar=\frac{4+2 \cos \theta}{2}, \quad k=\frac{3+\sqrt{2} \sin \theta}{2} \\
& (h-2)^2+\left(\frac{2 k-3}{\sqrt{2}}\right)^2=1 \\
& \therefore \quad 2\left(1-e^2\right)=1 \\
& 1-e^2=\frac{1}{2} \\
& e=\frac{1}{\sqrt{2}} \\
& \qquad \begin{array}{c}
1 \\
\text { Hence, the answer is } \frac{1}{\sqrt{2}}
\end{array}
\end{aligned}
$

Example 4: The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ is :
[JEE MAINS 2021]

Solution

Let us take any parametric point $B(2 \cos \theta, 3 \sin \theta)$ on the ellipse
Let $P_A$ be the mid-point of $A(-3,-5) \& B(h, k)$

So

$
\begin{aligned}
& h=\frac{2 \cos \theta-3}{2} \Rightarrow \cos \theta=\frac{2 h+3}{2} \\
& k=\frac{3 \sin \theta-5}{2} \Rightarrow \sin \theta=\frac{2 k+5}{3}
\end{aligned}
$
Square and add;

$
\cos ^2 \theta+\sin ^2 \theta=\left(\frac{2 h+3}{2}\right)^2+\left(\frac{2 k+5}{3}\right)^2=1
$
Replace $h \rightarrow x \& k \rightarrow y$

$
\begin{aligned}
& \Rightarrow \frac{4 x^2+12 x+9}{4}+\frac{4 y^2+20 y+25}{9}=1 \\
& \Rightarrow 36 x^2+108 x+81+16 y^2+80 y+100-36 \\
& \Rightarrow 36 x^2+16 y^2+108 x+80 y+145=0
\end{aligned}
$

Hence, the answer is $36 x^2+16 y^2+108 x+80 y+145=0$

Example 5: If the point P on the curve $4 x^2+5 y^2=20$ is farthest from the point $Q(0,-4)$, then $P Q^2$ is equal to:
[JEE MAINS 2020]

Solution:
Given ellipse is $\frac{x^2}{5}+\frac{y^2}{4}=1$

$
\begin{aligned}
& \text { Let point } P \text { is }(\sqrt{5} \cos \theta, 2 \sin \theta) \\
& (\mathrm{PQ})^2=5 \cos ^2 \theta+4(\sin \theta+2)^2 \\
& (\mathrm{PQ})^2=\cos ^2 \theta+16 \sin \theta+20 \\
& (\mathrm{PQ})^2=-\sin ^2 \theta+16 \sin \theta+21 \\
& =85-(\sin \theta-8)^2
\end{aligned}
$

will be maximum when $\sin \theta=1$

$
\Rightarrow(\mathrm{PQ})^2 \max =85-49=36
$
Hence, the answer is 36