# Examples on Ellipse

Example - Set I
1. For the following ellipse, find the lengths of major and minor axes, eccentricity coordinates of foci, vertices and equation of directrices.
9x2 + 25y2 = 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 b2 = a2 (1 âˆ’ e2) we get 9 = 25 (1 âˆ’ e2) or 25e2 = 16 or .
Since the denominator of x2 is larger than the denominator of y2, 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 .

1. For the following ellipse, find the lengths of major and minor axes, eccentricity, coordinates of foci, vertices and equation of directrices.
16x2 + 25y2 = 400
Solution
We divide the given equation by 400 obtain .
Since the denominator of x2 is larger than the denominator of y2 the major axis is along x-axis and the minor axis is along y-axis.

Also, a2 = 25 and b2 = 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 b2 = a2 (1 âˆ’ e2), we get 16 = 25 (1 âˆ’ e2)
â‡’ 25 e2 = 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 .

1. For the following ellipse, find the lengths of major and minor axes, eccentricity, coordinates of foci, vertices and equation of directrices.
3x2 + 2y2 = 6
Solution
Dividing the given equation by 6, we get
Since the denominator y2 is larger, the major axis is along y-axis and the minor axis is along x-axis.

Also, a2 = 3 and b2 = 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 b2 = a2 (1 âˆ’ e2), we get 2 = 3(1 âˆ’ e2) or 3e2 = 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
1. 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

1. 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

1. 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,     b2 = a2âˆ’ a2 e2 = 36 âˆ’ 25 = 11
so, an equation of the required ellipse

1. 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     ---------------(1)
Since it passes through (4, 3) and (-1, 4), we get and Multiplying the second equation by 16 and subtracting the first equation from it,We get
Thus,
Putting these values (1), we get an equation of the required ellipse as

1. 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  --------------- (1)
where a > b.
As it passes through (4, 1), we have
â‡’ 16(a2 âˆ’ 9) + a2 = a2(a2 âˆ’ 9) â‡’ a4 âˆ’ 26a2 + 144 = 0
â‡’ (a2 âˆ’ 18) (a2 âˆ’ 8) = 0 â‡’ a2 = 18 or a2 = 8.
When a2 = 18, b2 = a2 - 9 = 9, and when a2 = 8, b2 = a2 - 9 = âˆ’ 1.
As b2 cannot be negative, a2 = 8 is not possible.
Therefore a2 = 18 and b2 = 9.
Substituting these values in (1), we get an equation of the required ellipse as

1. 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 ------------------(1)
where a>b and b2 = a2 (1 - e2). As , we get .
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

1. 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 b2 = a2 (1 âˆ’ e2), we get

Substituting these values in the equation we get the equation of the required ellipse as

1. 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
where .
Thus, equation of the ellipse becomes ---------------(1)
Since this ellipse passes through (-3, 1), we must have

Substituting these values in (1), we get