Ellipse Parabola Hyperbola

An ellipse is a plane curve that results from the intersection of a cone by a plane in a way that produces a closed curve.

A parabola is a conic section, the intersection of a right circular conical surface and a plane to a generating straight line of that surface. Given a point and a corresponding line on the plane, the locus of points in that plane that are equidistant from them is a parabola.

A hyperbola is a curve, specifically a smooth curve that lies in a plane, which can be defined either by its geometric properties or by the kinds of equations for which it is the solution set.(Source: Wikipedia)



Example ellipse problem :

Find the equation of ellipse with its center at the origin and having foci at (`+- 2sqrt(6)`, 0 )and an eccentricity

equal to the  ,`(2sqrt(6))/(7)`

Solution:

The focal points on the X axis is the ellipse is oriented, view A, and the standard form of the equation is ` (x^2)/(a) + (y^2)/(b)`

The center of the origin, the numerators of the fractions on the left is x2 and y2. The problem is find the values of a and b.

The distance from the center to be either of the foci is equal to c. so in this problem

C = `2sqrt(6)`

from the given coordinates of the foci.

The values of a, c, and e (eccentricity) are related by

c=a*e

or

` a = (c)/(e)`

Substitute the values from c and e,

a = `(2sqrt(6))/((2sqrt(6))/(7))`

and

a = 7

a2= 49

Then, using the formula

b = `sqrt(a^2-c^2)`

or

b2 = a2 - c2

and substituting for a2 and c2,

b2 = 49 - (`2sqrt(6)` )2

b2= 49 - (4)(6)

b2= 49 - 24

gives the final required value of

b2 = 25

Then the equation of the ellipse is

`(x^2)/(49) + (y^2)/(25) ` = 1

Example parabola problem :

Reduce the equation y2-6y = 8x – 1 to standard form.

Solution:

Rearrange the equation of the 2nd -degree term and any 1st -degree terms of the same unknown are on the left side. Then group the unknown term appearing only in the 1st degree and all constants on the right:

y2 - 6y = 8x -1


Then complete the square in y:

y2 - 6y + 9 = 8x – 1 + 9

(y - 3)2 = 8x + 8

To get the equation in the form

(y - k)2 = 4a(x-h)

Factor an 8 out of the right side. Thus,

(y - 3)2= 8(x + 1)

is the equation of the parabola with its vertex at (-1,3).

Example hyperbola problem :

Find the equation of the hyperbola with an eccentricity of 3/2, directories x = ± 4/3, and foci at ( ± 3,0).

Solution:

The foci of the X axis at the points are (3,0) and ( - 3,0), so the equation is of the form ` (x^2)/(a^2)+(y^2)/(b^2)` = 1

First, the foci are given as (± 3,0); and since the foci are also the points (± c,0), then c=3

The eccentricity is given and the value of a2 can be determined from the formula

c = a * e

a = c/(e)

a = `(3)/((3)/(2))`

a =` (6)/(3)`

a = 2

a2 = 4

The relationship of hyperbola a, b and c are b2 = c2 – a2

And

b2 = (3)2 – (2)2

b2 = 9 – 4

b2 = 5

When these values are substituted in the equation `(x^2)/(a^2)+(y^2)/(b^2)` = 1 the equation `(x^2)/(4)+(y^2)/(5)` = 1

results and is the equation of the hyperbola.

The directories are given as x = `(4)/(3)` and, since

d = `(a)/(e)` or

a=de

Substituting the values given for d and e results in

a = `(4)/(3)((3)/(2))`

therefore,

a=2

and

a2 = 4

The value of c can be determined by the given equation,

c = ae

and a has been found to equal 2 and e is given as(3)/(2); therefore,

c =` 2((3)/(2))`

With the values of a and c computed, the value of b is found as before and the equation can be written.