Make Ellipse Equation

Under the picture that relates the foci, eccentricity, and length of the main and negligible axes. Derive the identities if we remember that if T is a point on the ellipse the addition of the distances from T to the two foci is a steady.  Derive the equation for the ellipse with vertical main axis.  The derivative for the horizontal main axis case is similar.

Vertical Ellipse and Horizondal Ellipse

Making ellipse equation:

Consider the upside of the ellipse.  The addition of the distances from the two foci to the up side of the make a ellipse is

(b –c) + (b + c) = 2b

The square of is given by

4b2

Then consider the right hand side of the ellipse.

See that the distances from each side to the right hand side is the equal.  The square of sum is

(2d)2 = 4d2

Using the Pythagorean Theorem,

d2 = a2 + c2

Replacement and setting these same to each other gives

4b2 = 4(a2 + c2)

So that

b2 = a2 + c2

Place the equation in the particularly simple form

`(x^2)/(a^2) + (y^2)/(b^2) =1 (or) (x^2)/(b^2) + (y^2)/(a^2) =1`

Make ellipse equations - Examples:

Make ellipse equations - Example1:

Given the following equation

9x2 + 4y2 = 36

a) Solve the x and y intercepts of the graph of the equation.

b) Solve the co-ordinates of the foci.

c) Solve the length of the main and slight axes.

d) Make the graph of the equation.

Solution:

9x2 / 36 + 4y2 / 36 = 1

x2 / 4 + y2 / 9 = 1

x2 / 22 + y2 / 32 = 1

a = 3 and b = 2 (NOTE: a >b) .

Set y = 0 in the equation get and Solve the xaxes

x2 / 22  = 1

Solve for x.

x2  = 22

x = ± 2

Set x = 0 in the equation get and Solve the y axes

y2 / 32 = 1

Solve for y.

y2  = 32

y = ± 3

b) We need to solve c first.

c2 = a2 - b2

a and b were found in part a).

c2 = 32 - 22

c2 = 5

Solve for c.

c = +- (5)1/2

The foci are    F1 (0, (5)1/2) and  F2 (0 , -(5)1/2)



c) The main axis length is given by  2 a = 6.

The slight axis length is given by  2 b = 4.

d) The graph of the equation is


Example Ellipse