Ellipse - Equation, Formula, Properties, Graphing

The four curves - circle, parabola, ellipse, and hyperbola are called conic sections because they can be formed by interesting a double right circular cone with a plane. 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 in race tracks, architectural design, mirrors, and celestial orbits.

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This Story also Contains

1. What is the Ellipse?
2. Standard Equation of Ellipse
3. Derivation of the equation of ellipse
4. Important Terms related to ellipse:
5. Eccentricity of the Ellipse
6. Ellipse Formulas
7. Solved Example Based on Ellipse

In this article, we will cover the concept of 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 twenty-five questions have been asked on JEE MAINS( 2013 to 2023) from this topic including five in 2020, four in 2021, and four in 2022.

What is the Ellipse?

An ellipse is the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).

OR

The locus of a point moves in a plane such that the ratio of the distance from a fixed point (focus) to the distance from a fixed line (directrix) is constant. The constant is known as eccentricity e and for ellipse 0 < e < 1.

Ellipse Shape

In geometry, an ellipse is a two-dimensional shape, that is defined along its axes. An ellipse is formed when a cone is intersected by a plane at an angle with respect to its base.

It has two focal points. The sum of the two distances to the focal point, for all the points in curve, is always constant.

A circle is also an ellipse, where the foci are at the same point, which is the center of the circle.

Major and Minor Axis

Ellipse is defined by its two-axis along x and y-axis:

• Major axis
• Minor Axis

We observe that the graph of the ellipse is a closed curve. The line that passes through the focus and is perpendicular to the directrix is called the major axis (focal axis) of the ellipse. In the figure, the major axis intersects the ellipse at points A and A' which are two vertices of the ellipse. Here, AA' is the major axis (or length of the major axis). It is the longest chord of the ellipse. The graph of an ellipse is symmetric about the major axis.

We have another axis of the ellipse called the minor axis which is the perpendicular bisector of AA'. The graph of the ellipse is symmetrical about the minor axis also. The minor axis meets the ellipse at points B and B'. BB' is the minor axis (or length of the minor axis). It is the shortest chord of the ellipse. Both axes intersect at point C, which is called the centre of the ellipse. The Centre bisects every chord of the ellipse passing through it.

Properties

• The ellipse has two focal points, also called foci.
• The fixed distance is called a directrix.
• The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
• The total sum of each distance from the locus of an ellipse to the two focal points is constant
• Ellipse has one major axis and one minor axis and a center

Standard Equation of Ellipse

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

Derivation of the equation of ellipse

Consider the figure, C is the origin, S is the focus and ZM is the directrix.
$SA=e.AZ$
$S{A}^{\prime }=$ e. $A^{\prime }Z$
(ii)
(from the definition)
Adding these
Let $A{A}^{\prime }=2a$. C is the midpoint of $A{A}^{\prime }$. Hence, $CA=C{A}^{\prime }=\mathrm{a}$.
$A=(a,0)$ and $A^{\prime }=(-a,0)$
From equation (i) and (ii)
$SA+S{A}^{\prime }=e(AZ+A^{\prime }Z)$
$2\mathrm{a}=\mathrm{e}(CZ-CA+C{A}^{\prime }+CZ)$
$2\mathrm{a}=\mathrm{e}(2\mathrm{CZ})$
$[CA=C{A}^{\prime }]$
$\mathrm{CZ}=\mathrm{a}/\mathrm{e}$
The equation of directrix, $ZM$ is $x=a/e$
Again from equation (i) and (ii)
$S{A}^{\prime }-SA=e(A^{\prime }Z-AZ)$
$[(C{A}^{\prime }+CS)-(CA-CS)]=e\left[A{A}^{\prime }\right]$
$\begin{array}{rl}& 2\mathrm{CS}=\mathrm{e}(2\mathrm{a})\\ & \mathrm{CS}=\mathrm{ae}\end{array}$
The coordinate of focus, $S=(\mathrm{ae},0)$ and $S^{\prime }=(-\mathrm{ae},0)$
$P(x,y)$ is any point on the ellipse and $PM$ is perpendicular to directrix $ZM$.

$\begin{array}{rl}& \frac{SP}{PM}=e\Rightarrow (SP{)}^2=e^2(PM{)}^2\\ & (x-ae{)}^2+(y-0{)}^2=e^2{(\frac{a}{e}-x)}^2\\ & x^2+a^2{e}^2-2aex+y^2=e^2{x}^2-2aex+a^2\\ & x^2(1-e^2)+y^2=a^2(1-e^2)\\ & \frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1\\ & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,b^2=a^2(1-e^2)\end{array}$

Important Terms related to ellipse:

1. Centre: All chord passing through point C is bisected at point C. Here C is the origin, i.e. (0, 0).

2. Foci: Point S and S’ are foci of the ellipse where, S is (ae, 0) and S’ is (-ae, 0).

3. Directrices: The straight-line ZM and Z’M’ are two directrices of the ellipse and their equations are x = a/e and x = -a/e.

4. Axis: In Figure AA’ is called the major axis and BB’ is called the minor axis. 2a is called the length of the major axis and 2b is called the length of the minor axis.

5. Double Ordinate: If a line perpendicular to the major axis meets the curve at P and P’, then PP’ is called double ordinate.

6. Latus rectum: Double ordinate passing through focus is called latus rectum. Here LL’ is a latus rectum. There is another latus rectum that passes through the other focus S'. So an ellipse has 2 latus rectum

Eccentricity of the Ellipse

The ratio of distances from the centre of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.

$e=\sqrt{1-\frac{b^2}{a^2}}$

Ellipse Formulas

An ellipse is a closed-shape structure in a two-dimensional plane. Hence, it covers a region in a 2D plane. So, this bounded region of the ellipse is its area.

Area of Ellipse

The area of the circle is calculated based on its radius, but the area of the ellipse depends on the length of the minor axis and major axis.

Area of the circle = πr²

And,

Area of the ellipse = π x Semi-Major Axis x Semi-Minor Axis

Perimeter of Ellipse

The perimeter of an ellipse is the total distance run by its outer boundary. For a circle, it is easy to find its circumference, since the distance from the centre to any point of the locus of the circle is the same. This distance is called the radius.

But in the case of an ellipse, we have two axes, major and minor, that cross through the centre and intersect. Hence, we use an approximation formula to find the perimeter of an ellipse, given by:

$p=2\pi \sqrt{\frac{a^2+b^2}{2}}$

Where a and b are the lengths of semi-major and semi-minor axes respectively.

Latus Rectum

The line segments perpendicular to the major axis through any of the foci such that their endpoints lie on the ellipse are defined as the latus rectum

The length of the latus rectum is 2b²/a.

L = 2b²/a

where a and b are the length of the minor axis and major axis.

Recommended Video Based on Ellipse

Solved Example Based on Ellipse

Example 1: Let ${\mathrm{PQ}}_{\text{be a }}$ a focal chord of the parabola ${\mathrm{y}}^2=4{\mathrm{x}}_{\text{such }}$ that it subtends an angle of $\bar{2}$ at the point $(3,0)$. Let the line segment $PQ$ be also a focal chord of
$\mathrm{E}:\frac{x^2}{{\mathrm{a}}^2}+\frac{y^2}{{\mathrm{b}}^2}=1,{\mathrm{a}}^2>{\mathrm{b}}^2$. If e is the eccentricity of the ellipse E , then the value of $\frac{1}{{\mathrm{e}}^2}$ is equal to :
[JEE MAINS 2022]
Solution

Slope of $\mathrm{PA}=\frac{2{\mathrm{t}}_1}{{\mathrm{t}}_1^2-3}$
Slope of QA $=\frac{2{\mathrm{t}}_2}{{\mathrm{t}}_2^2-3}$
Now these are perpendicular

$\begin{array}{rl}& \Rightarrow \frac{2t_1}{t_1^2-3}\cdot \frac{2t_2}{t_2^2-3}=-1\\ & \Rightarrow 4t_1{t}_2=-1(t_1^2-3)(t_2^2-3)\\ & \text{ Using }{t}_1{t}_2=-1\text{ for focal chord }\\ & \Rightarrow -4=-1(t_1^2-3)(\frac{1}{t_1^2}-3)\\ & \Rightarrow (t_1^2-3)(\frac{1}{t_1^2}-3)=4\\ & \Rightarrow 1-\frac{3}{t_1^2}-3t_1^2+9=4\\ & \Rightarrow 3t_1^2+\frac{3}{t_1^2}=6\end{array}$
$\begin{array}{rl}& \Rightarrow 3({\mathrm{t}}_1^2+\frac{1}{{\mathrm{t}}_1^2}-2)=0\\ & \Rightarrow {({\mathrm{t}}_1-\frac{1}{{\mathrm{t}}_1})}^2=0\\ & \Rightarrow {\mathrm{t}}_1=\frac{1}{{\mathrm{t}}_1}\\ & \Rightarrow {\mathrm{t}}_1^2=1\\ & \Rightarrow {\mathrm{t}}_1=1\text{ or }-1\\ & \therefore \mathrm{P}(1,2),\mathrm{Q}(1,-2)\end{array}$

$\therefore \mathrm{PQ}$ is also the focal chord and latus rectum of the ellipse
$\mathrm{P}(1,2)$ is end point of latus rectum
$\therefore$ ae $=1$ and $\frac{{\mathrm{b}}^2}{\mathrm{a}}=2$
$\mathrm{a}=\frac{1}{\mathrm{e}}$ and ${\mathrm{b}}^2=2\mathrm{a}$

Now $e^2=1-\frac{b^2}{a^2}$
$\begin{array}{rl}& {\mathrm{e}}^2=1-\frac{2\mathrm{a}}{{\mathrm{a}}^2}\\ & {\mathrm{e}}^2=1-\frac{2}{\mathrm{a}}\\ & {\mathrm{e}}^2=1-2\mathrm{e}\\ & {\mathrm{e}}^2+2\mathrm{e}-1=0\\ & \mathrm{e}=\frac{-2+\sqrt{4+4}}{2}\\ & =\frac{-2+2\sqrt{2}}{2}\\ & =\sqrt{2}-1\\ & \therefore \frac{1}{{\mathrm{e}}^2}=\frac{1}{(\sqrt{2}-1{)}^2}=\frac{1}{(3-2\sqrt{2})}\cdot \frac{(3+2\sqrt{2})}{(3+2\sqrt{2})}\\ & =3+2\sqrt{2}\end{array}$
Hence the correct answer is $3+2\sqrt{2}$

Example 2: If the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ meets the line $\frac{\mathrm{x}}{7}+\frac{\mathrm{y}}{2\sqrt{6}}=1\frac{\mathrm{x}}{7}-\frac{\mathrm{y}}{2\sqrt{6}}=1$ on the x -axis and the line y -axis, then the eccentricity of the ellipse is Solution: Line 1 and ellipse cut the x -axis at $\mathrm{x}=7$ and line 2 and ellipse cut the y -axis at $\mathrm{y}=2\sqrt{6}$.

$\begin{array}{rl}& \therefore \mathrm{a}=7,\mathrm{b}=2\sqrt{6}\\ & {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=1-\frac{24}{49}=\frac{25}{49}\\ & \Rightarrow \mathrm{e}=\frac{5}{7}\end{array}$
Hence, the answer is $\frac{5}{7}$

Example 3: Let the eccentricity of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$, be $\frac{1}{4}$. If this ellipse passes through the point $(-4\sqrt{\frac{2}{5}},3)$, then ${\mathrm{a}}^2+{\mathrm{b}}^2$ is equal to?
Solution

${\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{1}{16}\Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{15}{16}$
Also, an ellipse passes through $(-4\sqrt{\frac{2}{5}},3)$
So $\frac{32}{5{\mathrm{a}}^2}+\frac{9}{{\mathrm{b}}^2}=1$ Put ${\mathrm{b}}^2=\frac{15{\mathrm{a}}^2}{16}$

$\begin{array}{rl}& \Rightarrow \frac{32}{5{\mathrm{a}}^2}+\frac{9}{15{\mathrm{a}}^2}\times 16=1\Rightarrow \frac{80}{5{\mathrm{a}}^2}=1\Rightarrow {\mathrm{a}}^2=16,{\mathrm{b}}^2=15\\ & {\mathrm{a}}^2+{\mathrm{b}}^2=31\end{array}$
Hence, the answer is 31

Example 4: Let an ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a^2>b^2$, passes through $(\sqrt{\frac{3}{2}},1)$ and has eccentricity $\frac{1}{\sqrt{3}}$. If a circle, centered at focus $\mathrm{F}(\alpha ,0),\alpha >0$, of E and radius $\frac{2}{\sqrt{3}}$, intersects $E$ at two points $P$ and $Q$, then PQ is equal to [JEE MAINS 2021]
Solution

$\begin{array}{rl}& (\sqrt{\frac{3}{2}},1)\Rightarrow \frac{3}{2a^2}+\frac{1}{b^2}=1\Rightarrow \frac{3b^2}{2a^2}+1=b^2\\ & e=\frac{1}{\sqrt{3}}\Rightarrow \frac{1}{3}=1-\frac{b^2}{a^2}\Rightarrow \frac{b^2}{a^2}=\frac{2}{3}\\ & \text{ so }\frac{3}{2}\cdot \frac{2}{3}+1=b^2\Rightarrow b^2=2\Rightarrow a^2=3\\ & \text{ Focus }\to (ae,0)=(\sqrt{3}\cdot \frac{1}{\sqrt{3}},0)=(1,0)\\ & \therefore F(1,0)\Rightarrow \alpha =1\end{array}$

equation of circle : $(x-1{)}^2+y^2=\frac{4}{3}$
ellipse $:\frac{x^2}{3}+\frac{y^2}{2}=1$
Solving these $2\to P(1,\frac{2}{\sqrt{3}}),Q(1,\frac{-2}{\sqrt{3}})$

$P{Q}^2=\frac{16}{3}$
Hence, the answer is $\frac{16}{3}$

Example 5: If the points of intersection of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the circle $x^2+y^2=4b,b>4$ lie on the curve $y^2=3x^2$ then b is equal to :
[JEE MAINS 2021]
Solution

$y^2=3x^2$

and

$x^2+y^2=4b$
Solve both we get $x^2=b$
Now $\frac{x^2}{16}+\frac{y^2}{b^2}=1$
so $\frac{x^2}{16}+\frac{3x^2}{b^2}=1$

$\begin{array}{rl}& \frac{b}{16}+\frac{3}{b}=1\\ & b^2-16b+48=0\\ & (b-12)(b-4)=0\\ & b=12,b>4\end{array}$

Hence, the answer is 12