Parametric Equation of an Ellipse

The ellipse is the locus of a point such that the ratio of its distance from a fixed point (focus) and a fixed line (directrix) is constant, the value of which is always less than 1. The constant ratio is called eccentricity e. In real life, we use Ellipse Airplane wings, rudders, and shapes of boat keels.

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This Story also Contains

1. Standard Equation of Ellipse
2. What is the Parametric equation of Ellipse?
3. Derivation of Parametric Equation of Ellipse
4. What is Auxillary Circle?
5. Summary
6. Solved Examples Based on the Parametric equation of Ellipse

In this article, we will cover the concept of the Parametric equation of the Ellipse. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of nineteen questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2017, one in 2020, one in 2021, one in 2022, and two in 2023.

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