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# Eccentricity of Ellipse

Eccentricity in mathematics denoted by "e" is associated with conics. It can be taken as a measure of deviation of the conic section from a circular section.

If we have to write this in terms of definition of conic, which states that a conic section is defined as the locus of points whose distance from a fixed point (focus) and a fixed line (directrix) remains a constant. Ratio of the distance between the fixed point and the point and the point and the directrix is known as the eccentricity.
• Eccentricity of circle is zero (e=0).
• Eccentricity of a Ellipse is greater than zero and lesser than 1 (0<e<1)
• Eccentricity of a parabola  is 1 (e=1)
• Eccentricity of a Hyperbola is greater than one (e>1)

Formula for eccentricity

First let us see a standard formula for ellipse with the center at origin

$\frac{x^2}{a^2}$ + $\frac{y^2}{b^2}$ = 1

In this $a$ is the length of the semi major axis, that is the distance from center to the vertex on the major axis and $b$ is the length of the semi minor axis which is the distance from the center to the vertex on the minor axis.

The eccentricity formula for ellipse is given in terms of these $a$ and $b$

$e =$  $\sqrt{(1 - \frac{b^2}{a^2})}$

We can also write the formula in terms of "c" which gives the length from the center to the focus.

The formula for c is c = $\sqrt{(a^2-b^2)}$

Using this we can get the formula for eccentricity of ellipse

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

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

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

e = $\sqrt{\frac{a^2-b^2 }{a}}$

e = $\frac{c}{a}$

Examples

Example 1:  For the ellipse $\frac{x^2}{16}$ + $\frac{y^2}{9}$ = 1 find the length of major axis, minor axis and the eccentricity.

Solution:  The given ellipse is $\frac{x^2}{16}$ + $\frac{y^2}{9}$ = 1 comparing with the standard form $\frac{x^2}{a^2}$ + $\frac{y^2}{b^2}$ = 1  we get a=4 and b = 3, since a>b the ellipse is horizontal.

The length of the major axis = 2a

The length of the major axis = 2(4) = 8

The length of the minor axis = 2b

The length of the minor axis = 2(3) = 6

The eccentricity of the ellipse is the given by the formula

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

e = $\sqrt{(1 - \frac{3^2}{4^2})}$

e = $\sqrt{\frac{(4^2-3^2)}{4^2}}$

e = $\sqrt{\frac{(16-9)}{4}}$

e = $\frac{\sqrt 7}{4}$

The eccentricity of this ellipse is $\frac{\sqrt 7}{4}$ this is approximately equal to 0.66 and we can see that it lies between 0 and 1.

Example 2:  Find the equation of the ellipse whose major axis is horizontal and passes through the point (2,1) and eccentricity = $\frac{1}{2}$

Solution:

We know the general equation of ellipse $\frac{x^2}{a^2}$ + $\frac{y^2}{b^2}$ = 1

It passes through (2,1) then we get $\frac{4}{a^2}$ + $\frac{1}{b^2}$ = 1....(1)

We know e = $\frac{1}{2}$

Using formula   e = $\sqrt{(1- \frac{b^2}{a^2})}$

Squaring

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

Substituting the value for e we get $\frac{1}{4}$ = 1- $\frac{b^2}{a^2}$.

We subtract 1 on both the sides we get $\frac{-3}{4}$ = $\frac{-b^2}{a^2}$

Simplifying we get $3a^2$ = $4b^2$ => $a^2$ = $\frac{4}{3}$ $b^2$

Replacing and solving (1) we get $\frac{3}{b^2}$ + $\frac{1}{b^2}$ = 1 => $\frac{4}{b^2}$ = 1 => $b^2$ = 4 => $a^2$ = $\frac{16}{3}$.  The equation is $\frac{(3x^2)}{16}$ + $\frac{y^2}{4}$ = 1

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