Equation of an ellipse in its standard form
x^{2}/a^{2} + y^{2}/b^{2} =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 A^{1} are points such that their distances from the focus bear a constant ratio e to their respective distances from the directrix. Therefore, A, A^{1} lie on the ellipse.
To trace the ellipse use the following points.
 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.
 Origin:
The curve does not pass through the origin.
 Intersection of the axes:
The curve meets x axis at y = 0. Putting y = 0 in equation (3), we get
 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 S^{1} as the focus and Z^{1}K^{1} as directrix, we have a second focus S^{1}(ae, 0) and a second directrix x = a/e.
Parts of the ellipse x^{2}/a^{2} + y^{2}/b^{2} = 1 a>b
Vertices:
The points A and A^{1} 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 S^{1}(ae,0) are called the foci of the ellipse.
Major and Minor Axes:
The distances AA^{1} = 2a and BB^{1} = 2b are called the major and minor axes, respectively.
Since e < 1 a>b
AA^{1} > BB^{1}
Directrices:
ZK and Z^{1}K^{1} 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 AA^{1} and BB^{1}.
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 semilatus rectum.
The coordinates of L are (ae,SL)
L lies on x^{2}/a^{2} + y^{2}/b^{2} = 1
Focal distances of a point on an ellipse
If P(x,y) is a point on the ellipse, then the distances SP and S^{1}P are the focal distances.
SP = ePM = eNK = e(CKCN) = e(a/ex) = aex
S^{1}P = ePM^{1} = eNK1 = e(CK^{1}CN) = e(a/e+x) = a+ex
5.4.6 Equation of an ellipse in another form
The ellipse given by x^{2}/a^{2} + y^{2}/ b^{2} = 1 a>b has its major and minor axes on the xaxis and yaxis, respectively.
If, however, x^{2}/a^{2} + y^{2}/b^{2} = 1 a<b, then BB^{1} becomes the major axis and AA^{1} the minor axis.
The coordinates of the foci are (0,be), (0,be), the equations to the directories. ZK and Z^{1}K^{1} are y = ± b/e, where e is the eccentricity given by the formula.
Equation of the ellipse 
x^{2}/a^{2} + y^{2}/b^{2} = 1 a>b 
x^{2}/a^{2} + y^{2}/b^{2} = 1 a<b 
Coordinates of the center 
(0,0) 
(0,0) 
Coordinates of the vertices 
(a,0), (a,0) 
(0,b), (0,b) 
Coordinates of the foci 
(ae,0), (ae,0) 
(0,be), (0,be) 
Length of the major axis 
2a 
2b 
Length of the minor axis 
2b 
2a 
Equation of the major axis 
y = 0 
x = 0 
Equation of the minor axis 
x = 0 
y = 0 
Equations of the directrices 
x = a/e x = a/e 
y = b/e y= b/e 
Eccentricity 


Length of the latus rectum 
2b^{2}/a 
2a^{2}/b 
Focal distance of a point (x,y) 
a ± ex 
b± ey 
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 xy+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 crossmultiplying
8(x^{2} + y^{2} + 2x  2y + 2) = x^{2} + y2 + 9  2xy + 6x  6y
8x^{2} + 8y^{2} + 16x  16y  x^{2}  y^{2}  9 + 2xy  6x + 6y = 0
7x^{2} + 7y^{2} + 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.
 16x^{2} + 25y^{2} = 400
 3x^{2} + 2y^{2} = 6
 x^{2} + 4y^{2}  2x = 0
Solution:
 Major axis is along the xaxis
Minor axis is along the yaxis
Length of the major axis = AA^{1} = 2a = 2 * 5 = 10
Length of the minor axis = BB^{1} = 2b = 2 * 4= 8
Vertices:
Vertices are (a, 0), (a, 0)
or (5, 0), (5, 0)
 Eccentricity
 = 3/5
 Foci:
Coordinates are (ae,0), (ae,0)

Eccentricity:
 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 xaxis
Minor axis is along the yaxis
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:
 Find the center, foci, vertices, eccentricity, length of axis and directrices of the following ellipses. Draw a rough sketch.
 X^{2} + 4y^{2}  4x + 24y + 31 = 0
 3x^{2} + 4y^{2}  12x  8y + 4 = 0
 16x^{2} + 4y^{2}  32x  24y  12 = 0
 Show that x^{2} + 4y^{2} + 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.
Answers:
 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 x^{2}/ a^{2} + y^{2}/ b^{2} = 1
Eccentricity:
 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 yaxis with respect to the new axis are
=  .7
Points are (2, 2.7), (2,  .7)
The figure of the ellipse is as follows.
 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
X^{2} / (2)^{2} + Y^{2} / (4)^{2} =1
This is of the form X^{2}/a^{2} + Y^{2}/b^{2} = 1
where a^{2} = (2)^{2} a = 2
 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 xaxis
Minor axis is the yaxis
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