# Finding The Equation Of An Ellipse

**Finding the Equation of an Ellipse **

An ellipse is the set of all points in a plane such that the sum of the distances of each point from two fixed points is the same, The fixed points are called the “foci” and the line through the foci is called the axis of symmetry. The ratio c/a is called the eccentricity of the ellipse. The point(h, k) that is midway between the foci is called the center of the ellipse.The intersection points of ellipse and the line through the foci are known as the vertices. The line segment joining the vertices is the major axis . The part of the line through the center perpendicular to the major axis and intercepted by the ellipse is the minor axis.

Stanndard Forms of the Equation of an Ellipse

(1) (X – h )^2 / a^2 + (Y – k )^2/b^2 = 1

Is an equation of the ellipse with center at (h, k), vertices at (h+- a, k ) ,foci at

(h +- c, k), major axis parallel to the X-axis and semi axes of length a and b.

(2) (Y – k )^2/a^2 + (X – h)^2/b^2 = 1

Is an equation of ellipse with center at (h, k), vertices at (h, k +- a), foci at

(h, k +- c), major axis parallel to the Y-axis and semiaxes of length a and b.

If the center is at the origin or (0,0), the equations above have the form :

X^2/a^2 + Y^2/b^2 = 1 and

Y^2/a^2 + X^2/b^2 = 1

In each case a^2 = b^2 + c^2 and thus a > b and a > c.

Also for every ellipse a is the distance from center to vertex and c is the distance from center to focus.

**Problem Number One :**

**Find the equation of the ellipse with center at (-1. 1) and a focus at (3, 1) and a vertex at (4, 1).**

Solution :

We can write the equation of an ellipse if we know the center, the semiaxes and which of the two standard forms to use. With the given data, the center, focus and vertex are on the line parallel to the X-axis therefore the major axis of the ellipse is parallel to the X-axis. We will use the standard equation (1). Since a is the distance between the center and the vertex, we solve for a by just simply getting the directed distance between the X-coordinates of the center and the vertex.

a = 4 – (-1) ==> a = 4 + 1 è. a = 5

Also, since c is the distance between the center and the focus we solve for c by simply getting the directed distance between the X-coordinates of the center and the focus.

c = 3 – (-1) è c = 3 + 1 c = 4

We now solve for b using the formula a^2 = b^2 + c^2

b = SQRT(a^2 – c^2)

b = SQRT(25 -16) = 3

Since now we know the values of a and b we just simply substitute these values to standard equation (1) of an ellipse.Thus

(X + 1)^2/25 + (Y – 1)^2/9 = 1

is the equation of the ellipse we are solving.

Problem Number Two :

Find the equation of the ellipse whose ends of the minor axis are at (-4, 3) and (6,3)

and a vertex at (1, 10).

Solution :

The midpoint of the minor xix is the center of the ellipse.

X = ( X1 + X2)/2 = (- 4+ 6)/2 = 2/2 = 1

Therefore (1, 3 ) is the center of the ellipse whose equation we are solving for.

Since the minor axis is parallel to the X-axis, the major axis of the ellipse we are solving is parallel to the Y-axis.

In order to get a , get the directed distance between the vertex (1,10) and the center (1, 3). Just get the directed distance between the Y-coordinates of the center and vertex.

a = 10 – 3 = 7.

To get b, get the directed distance between the X-coordinates (1, 3 ) and one endpoint of the minor axis (6, 3);

b = 6 – 1 = 5.

Now we have value for a and b and we have known the center we can now find the standard equation of the ellipse we are solving by simply substituting the designated values to standard equation number (2).

( Y – 3)^2/49 + (X - 1)^2/25 = 1

Problem Number Three :

Find the equation of the set of all points such that the sum of the distances of each point from (2, 5) and (2, -3) is 14.

Solution :

By the definition of the ellipse the given points are the foci, 2a is 14 and a = 7. The center is the midpoint of the line joining the foci namely (2, 1). Since 2c = 5 – (-3) = 8

We have c = 4 . Thus b^2 = 7^2 – 4^2 = 49 – 16 = 33 and the equation we are looking for is :

(Y – 1)^2/49 + (X – 2)^2/33 = 1

Problem Number Four :

Put the following equation in standard form and find the center, vertices and foci.

9X^2 + 4Y^2 - 36X – 8Y + 4 = 0

Group and Factor :

(9X^2 - 36X) + ( 4Y^2 - 8Y ) = -4

9(X^2 - 4X) + 4( Y^2 - 2Y ) = -4

Complete the square :

9 (X^2 -4X + 4 ) + 4(Y^2 – 2Y + 1) = -4 +34 + 4

9(X^2 – 4X + 4 ) + 4( Y^2 – 2Y + 1 ) = 36

Factor as Perfect Square Trinomial and divide the whole equation by 36 :

9(X – 2)^2/36 + 4(Y – 1)^2/36 = 36/36

(X – 2)^2/4 + (Y – 1)^2/9 = 1

Center is at (2, 1)

a = 3 and b = 2

c = SQRT(9 – 4 )

c = SQRT (5)

The major axis of the ellipse is parallel to the Y-axis

Vertices are at (2, 4) and (2, -3)

Foci are at ( 2, 1 + SQRT(5)) and (2, 1 – SQRT(5) .

SOURCE:

COLLEGE ALGEBRA

By

Rees

Sparks

Rees

## Comments

Hey, these hubs are nice. Wish I found them 4 years back! But I think you are not making full use of the text-editor features when writing math expressions. Like there is a superscript button to write x-square (x^2)

Sorry Ate Cristina but I haven't got a clue what it is this is all about. I know it is Math but I don't understand one bit of it. I guess past simple arithmetic, I am math illiterate.

KUYA Dave.

What a great and novel way for someone to study for an Algebra II test on conic sections. Surely, if you can write a quality hub about the topic, then you have mastered the subject!

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