# Examples on Ellipse

**Example - Set I**

**For the following ellipse, find the lengths of major and minor axes, eccentricity coordinates of foci, vertices and equation of directrices.**

**9x**^{2 }+ 25y^{2 }= 225

**Solution**

We convert the given equation in standard form by dividing it with 225. The given equation becomes

This shows that a = 5, b = 3.

Substituting these values in the relation b

^{2 }= a

^{2}(1 âˆ’ e

^{2}) we get 9 = 25 (1 âˆ’ e

^{2}) or 25e

^{2 }= 16 or .

Since the denominator of x

^{2}is larger than the denominator of y

^{2}, the major axis is along x-axis, minor axis is along y-axis.

The foci are (ae, 0) = ( 4, 0) and the vertices are ( a, 0) = ( 5, 0).

Length of the major axis is 2a = 10 and that of minor axis is 2b = 6.

The equations of two directrices are .

**For the following ellipse, find the lengths of major and minor axes, eccentricity, coordinates of foci, vertices and equation of directrices.**

**16x**^{2 }+ 25y^{2 }= 400

**Solution**

We divide the given equation by 400 obtain .

Since the denominator of x

^{2}is larger than the denominator of y

^{2}the major axis is along x-axis and the minor axis is along y-axis.

Also, a

^{2 }= 25 and b

^{2 }= 16. Therefore length of the major axis is 10 and that of the minor axis is 8.

Substituting the values of a and b in the relation b

^{2 }= a

^{2}(1 âˆ’ e

^{2}), we get 16 = 25 (1 âˆ’ e

^{2})

â‡’ 25 e

^{2 }= 25 - 16 = 9 or or .

Foci of the ellipse are (ae, 0) = ( 3, 0) and the vertices are ( a, 0) = ( 5, 0).

The equations of two directrices or .

**For the following ellipse, find the lengths of major and minor axes, eccentricity, coordinates of foci, vertices and equation of directrices.**

**3x**^{2 }+ 2y^{2 }= 6

**Solution**

Dividing the given equation by 6, we get

Since the denominator y

^{2}is larger, the major axis is along y-axis and the minor axis is along x-axis.

Also, a

^{2 }= 3 and b

^{2 }= 2. Therefore, the length of the major axis is and that of the minor axis is. Substituting the values of a and b in the relation b

^{2}= a

^{2}(1 âˆ’ e

^{2}), we get 2 = 3(1 âˆ’ e

^{2}) or 3e

^{2 }= 1 or .

âˆ´ Foci of the ellipse are (0, Â± ae) = (0, Â± 1).

Vertices of the ellipse are (0, Â± a) = (0, Â± ).

Equation of Directrices are y = Â± 3

**Example - Set II**

**Find the equation of the ellipse satisfying the given condition.**

**Vertices at (****5, 0), foci at (4, 0)**

**Solution**

Since the vertices are ( 5, 0) and foci at ( 4, 0), the equation of the major axis is y = 0, i.e. x-axis. [Recall the major axis contains both the foci and both the vertices.]

Since, centre of the ellipse is the mid-point of the vertices (or foci), it is (0, 0), the origin in the present case. Therefore the minor axis (the one passing through the centre and perpendicular to the major axis) is x = 0, i.e. y-axis.

Therefore, an equation if the ellipse must be of the form ----------------(1)

We have a = 5, ae = 4, therefore . Also .

Substituting these values in (1), we get the equation of required ellipse as

**Find the equation of the ellipse satisfying the given condition**-

**Foci at (0, 8),**

**Solution**

In this case, the major axis is x = 0, i.e. y-axis, centre is (0, 0) and the minor axis is x-axis. We have , so that a=10.

Also .

We get a required equation as or

**Find the equation of the ellipse satisfying the given condition.**

**Foci at (****5, 0) and as one directrix.**

**Solution**

In this case, the major axis is x = 0, i.e. y-axis, centre is (0, 0) and the minor axis is x-axis. We are given ae = 5 and .

Therefore, .

Also, b

^{2}= a

^{2}âˆ’ a

^{2}e

^{2}= 36 âˆ’ 25 = 11

so, an equation of the required ellipse

**Find the equation of the ellipse satisfying the given condition.**

**Axes along the coordinate axes, passing through (4, 3) and (-1, 4).**

**Solution**

Let the equation of the ellipse be

**Find the equation of the ellipse satisfying the given condition.**

**Foci at (****3, 0) passing through (4, 1)**

**Solution**

In this case, major axis is y = 0, i.e. x-axis, centre is (0, 0) and the minor axis is y-axis. We are given that ae = 3

Let an equation of the required ellipse be

â‡’ 16(a

^{2 âˆ’ }9) + a

^{2 }= a

^{2}(a

^{2 âˆ’ }9) â‡’ a

^{4 âˆ’ }26a

^{2 }+ 144 = 0

â‡’ (a

^{2 âˆ’ }18) (a

^{2 âˆ’ }8) = 0 â‡’ a

^{2 }= 18 or a

^{2 }= 8.

When a

^{2 }= 18, b

^{2 }= a

^{2 }- 9 = 9, and when a

^{2 }= 8, b

^{2 }= a

^{2 }- 9 = âˆ’ 1.

As b

^{2}cannot be negative, a

^{2 }= 8 is not possible.

Therefore a

^{2 }= 18 and b

^{2 }= 9.

Substituting these values in (1), we get an equation of the required ellipse as

**Find the equation of the ellipse satisfying the given condition.**

**Eccentricity is and passing through (6, 4),**

**Solution**

Let an equation of the required ellipse be

Substituting these values in (1), we get

As this ellipse passes through (6, 4), we get

Therefore, b2 = 43.

Substituting these values in (1), we get equation of the required ellipse as

**Find the equation of the ellipse satisfying the given condition.**

**Axes along coordinate axes, vertex at (0, 7) and y = 12 as one directrix.**

**Solution**

As the vertex (0,7) lies on the y-axis, the major axis is along y-axis.

We have a = 7 and . Since b

^{2 }= a

^{2 }(1 âˆ’ e

^{2}), we get

Substituting these values in the equation

**Find the equation of the ellipse satisfying the given condition.**

**Eccentricity is and passing through (âˆ’ 3, 1).**

**Solution**

Let an equation of the ellipse be

Substituting these values in (1), we get