## Ellipse Equation

#### Equation of an ellipse in its standard form

x2/a2 + y2/b2 =1

Let S be the focus, ZK the directrix and e the eccentricity of the ellipse whose equation is required.

Draw SK perpendicular to ZK from S. Divide SK internally and externally at A and A1 (A1 lies on KS produced) respectively in the ratio e:1. The points A and A1 are points such that their distances from the focus bear a constant ratio e to their respective distances from the directrix. Therefore, A, A1 lie on the ellipse.    To trace the ellipse use the following points.

1. Symmetry
For every value of x, there are equal and opposite values of y (from equation (2)), and for every value of y there are equal and opposite values of x (from equation (3)).
So the curve is symmetric about both the axes.
2. Origin:
The curve does not pass through the origin.
3. Intersection of the axes:
The curve meets x axis at y = 0. Putting y = 0 in equation (3), we get 4. Region If x > a or x < -a then a2 - x2 < 0, we get imaginary values for y. Therefore, no part of the curve exists to the left of A1 and to the right of A.
If y > b or y < -b
then x has imaginary values and we get no part of the curve below B1 and above B.
From equation (2), we find that at x = 0, y = ± b and as the x values increase, the y values decrease so that y= 0 at x=a.
Therefore, the curve is a closed curve.
From equation (2), we find that at x = 0, y = ± b and as the x values increase, the y values decrease so that y= 0 at x=a.
Therefore, the curve is a closed curve.

#### Second focus and second directrix of an ellipse

In the figure of the ellipse given earlier, P(x,y) is a point on the curve and we have  This is the condition for the same ellipse. Since we started with S1 as the focus and Z1K1 as directrix, we have a second focus S1(-ae, 0) and a second directrix x = -a/e.

#### Parts of the ellipse x2/a2 + y2/b2 = 1 a>b

Vertices:

The points A and A1 where the ellipse meets the line joining the foci are called the vertices of the ellipse, and are (a,0) and (-a,0).

Foci:

The points S(ae,0) and S1(-ae,0) are called the foci of the ellipse.

Major and Minor Axes:

The distances AA1 = 2a and BB1 = 2b are called the major and minor axes, respectively.

Since e < 1 a>b

AA1 > BB1

Directrices:

ZK and Z1K1 are the two directrices of the ellipse and their equations are x = a/e and x = -a/e, respectively.

Center:

The center of a conic section is a point that bisects every chord passing through it. In the ellipse, every chord passing through C is bisected at C(0,0).

C is the center of the ellipse. C is the midpoint of AA1 and BB1.

Eccentricity: Ordinate and Double Ordinate

Let P be a point on the ellipse and let PN be perpendicular to the major axis AA1 such that PN meets the ellipse at P1. Then, PN is called the ordinate of P and PNP1 is called a double ordinate.

Latus rectum:

It is a double ordinate passing through the focus.

In the figure, LSL1 is the latus rectum, LS is called the semi-latus rectum.

The coordinates of L are (ae,SL)

L lies on x2/a2 + y2/b2 = 1 #### Focal distances of a point on an ellipse

If P(x,y) is a point on the ellipse, then the distances SP and S1P are the focal distances.

SP = ePM = eNK = e(CK-CN) = e(a/e-x) = a-ex

S1P = ePM1 = eNK1 = e(CK1-CN) = e(a/e+x) = a+ex

#### 5.4.6 Equation of an ellipse in another form

The ellipse given by x2/a2 + y2/ b2 = 1 a>b has its major and minor axes on the x-axis and y-axis, respectively.

If, however, x2/a2 + y2/b2 = 1 a<b, then BB1 becomes the major axis and AA1 the minor axis.

The coordinates of the foci are (0,be), (0,-be), the equations to the directories. ZK and Z1K1 are y = ± b/e, where e is the eccentricity given by the formula. Equation of the ellipse x2/a2 + y2/b2 = 1 a>b x2/a2 + y2/b2 = 1 a

#### Special form of an ellipse

If the center of the ellipse is at a point (h,k) and the directrices of the axes are parallel to the coordinate axes, then its equation is We shift the origin (0,0) to the point (h,k).

Example 2

Find the equation of an ellipse whose focus is (-1,1), the directrix is x-y+3=0 and the eccentricity is 1/2.

Let P(x,y) be a point on the ellipse. If PM = perpendicular from P, the directrix is By cross-multiplying

8(x2 + y2 + 2x - 2y + 2) = x2 + y2 + 9 - 2xy + 6x - 6y
8x2 + 8y2 + 16x - 16y - x2 - y2 - 9 + 2xy - 6x + 6y = 0
7x2 + 7y2 + 2xy + 10x - 10y + 7 = 0

Example 3

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

1. 16x2 + 25y2 = 400
2. 3x2 + 2y2 = 6
3. x2 + 4y2 - 2x = 0

Solution:

1. 2. 3. Major axis is along the x-axis
Minor axis is along the y-axis
Length of the major axis = AA1 = 2a = 2 * 5 = 10
Length of the minor axis = BB1 = 2b = 2 * 4= 8
Vertices:
Vertices are (a, 0), (-a, 0)
or (5, 0), (-5, 0)
4. Eccentricity 5. 6. 7. = 3/5
8. Foci:
Coordinates are (ae,0), (-ae,0) 9. 10. 11. Eccentricity: 12. Solution: the origin (0,0) becomes (1,0) as h=1, k=0.
Shifting the origin to (1,0), we obtain the new coordinates X,Y as
x - 1 = x                       x = x + 1
y - 0 = y            or         y = y
Equation (1) reduces to Major axis is along the x-axis
Minor axis is along the y-axis
Length of the major axis = 2a = 2 * 1= 2
Length of the minor axis = 2b = 2 * 1/2 = 1
Eccentricity: Vertices:
Coordinates of vertices with respect to the new axes are (a, 0), (-a, 0)
(X, 0) = (1, 0), (-X, 0) = (-1, 0)
With respect to old axes
x = X + 1
= 1 + 1
= 2
x = -X + 1
= -1 + 1
= 0
(2,0), (0,0)
Foci: Example 4:

Find the eccentricity, center, vertices, foci, minor axis, major axis, directrices and latus rectum of the ellipse

Solution:   Eccentricity: Center:

Coordinates with respect to the new axes (X,Y) = (0,0)

With respect to the old axes

x = X + 3

= 0 + 3

= 3

y = Y + 5

= 0 + 5

= 5

Center (3,5)

Vertices:

Coordinates with respect to the new axes are (0,b), (0,-b) = (0,5), (0,-5)

With respect to the old axes

x = X + 3

x = (0+3)

= 3

x = (0+3)

= 3

y = 5 + 5

= 10

y = - 5 + 5

= 0

Vertices are (3,10), (3,0)

Foci:

With respect to the new axes foci are (0,be), (0,-be) or

(0, 5 *4/5), (0, -5 *4/5)

(0,4), (0,-4)

With respect to the old axes

x = X + 3

= 0 + 3

= 3

x = X + 3

= 0 + 3

= 3

y = Y + 5

y = 4 + 5

= 9

y = Y + 5

= -4 + 5

= 1

Foci are (3,9), (3,1)

Directrices:

With respect to the new axes, the equations are With respect to the old axes

 y = Y + 5    = 25/4 + 5 y = 25 + 20/4 y = 45/4 or 4y - 45 = 0 y = Y + 5    = - 25/4 + 5 y = -25+20/4 y = -5/4 4y + 5 = 0

Length of the major axis = 2b = 2 * 5 = 10

Length of the minor axis = 2a = 2 *3 = 6

Equation of the major axis with respect to the new coordinates is X = 0

With respect to the old axes is

x = X + 3

x = 0 + 3

x = 3

Equation of the minor axis with respect to the new coordinates is Y = 0

With respect to the old axes is

y = Y + 5

y = 0 + 5

y = 5

Latus rectum:

The length of the latus rectum Equation of latus recta with respect to the new axes With respect to the old axes

y = Y + 5

= 4 + 5

y = 9

y = Y + 5

= -4 + 5

y = 1

The figure of the ellipse is given below. #### Try these questions:

1. Find the center, foci, vertices, eccentricity, length of axis and directrices of the following ellipses. Draw a rough sketch.
1. X2 + 4y2 - 4x + 24y + 31 = 0
2. 3x2 + 4y2 - 12x - 8y + 4 = 0
3. 16x2 + 4y2 - 32x - 24y - 12 = 0
2. Show that x2 + 4y2 + 2x + 16y + 13 = 0 is the equation to an ellipse. Find its eccentricity, center, foci, vertices, directrices, length of the latus rectum and the equations of the latus recta.
1. Solution: Shifting the origin to (-2, -3) and denoting the new coordinates with respect to the new axes, as X and Y which is of the form x2/ a2 + y2/ b2 = 1
2. Eccentricity: 3. Solution: Shifting the origin to (2, 1) and denoting the new coordinates with respect to the new axes as X, Y
x - 2 = X
x = X + 2
y - 1 = Y
y = Y + 1
Equation (1) now becomes Eccentricity: Foci:
With respect to the new axes the foci are (ae, 0), (-ae, 0)
= (2 *1/2, 0), (-2 *1/2, 0)
= (1, 0), (-1, 0)
With respect to the old axes
x = X + 2
= 1 + 2
= 3
y = Y + 1
= 0 + 1
= 1
and
x = X + 2
= - 1 + 2 = 1
y = Y + 1
= 0 + 1
= 1
Foci are (3, 1), (1,1)
Vertices:
With respect to the new axes are
(a, 0), (-a, 0)
= (2, 0), (-2, 0)
With respect to the old axes
x = X + 2
= 2+ 2
= 4
y = Y + 1
= 0 + 1
= 1
x = X + 2
= - 2+ 2
= 0
y = Y + 1
= 0 + 1
= 1
Vertices are ( 4, 1), (0, 1)
Center:
With respect to the new axes
(X, Y) = (0, 0)
With respect to the old axes
x = X + 2
= 0 + 2
= 2
y = Y + 1
= 0 + 1
= 1
Center (2, 1) Directrices:
With respect to the new axes
X = a/e                 X = -a/e
X = 2/(1/2)              X = -2/(1/2)
X = 4                    X = -4
X - 4 = 0               X + 4 = 0
With respect to the old axes
x = X + 2       and     x = X+ 2
x = 4 + 2       and     x = -4 + 2
x = 6       and     x = - 2
x - 6 = 0 and x + 2 = 0 are the directrices.
To find the values of the points where the ellipse cuts the minor axis or y-axis with respect to the new axis are = - .7
Points are (2, 2.7), (2, - .7)
The figure of the ellipse is as follows. 4. Solution: Shifting the origin to the point (1,3) and denoting the new coordinates with respect to the new axes as (X, Y)
x - 1= X
x = X + 1
and
y - 3= Y
y = Y + 3
Equation (1) now becomes
X2 / (2)2 + Y2 / (4)2 =1
This is of the form X2/a2 + Y2/b2 = 1
where a2 = (2)2             a = 2
3. Solution:  Shifting the origin to (-1, -2) and denoting the new coordinates with respect to the new axes as X,Y.
x + 1 = X
x = X - 1
y + 2 = Y
y = Y - 2
Equation (1) becomes Major axis is the x-axis
Minor axis is the y-axis
Center: With respect to the new axis ( 0, 0)
With respect to the old axis
x = X - 1
= 0 - 1
= -1
y = Y - 2
= 0 - 2
= -2
Center (-1, -2)
Eccentricity Vertices:
With respect to the new axes ( a, 0), (-a, 0)
= (2, 0), (-2, 0)
With respect to the old axes.
x = X - 1
= 2 - 1
= 1
y = Y - 2
= 0 - 2
= - 2
and
x = X - 1
= -2 - 1
= -3
y = Y - 2
= 0 - 2
= -2
Vertices are (1, -2), (-3, -2)
Foci:
With respect to the new axis
(ae, 0), (-ae, 0)
or Directrices:
With respect to the new axes  