Equation of an ellipse given c and eccentricity.

[bnosam]

Homework Statement

Write the equation of the conic that meets the conditions:

An ellipse that has the centre at (0, 0), has a horizontal major axis, the eccentricity is 1/2 and 2c = 1.

Homework Equations

[tex] \frac{(x - h)^2} {a^2} + \frac{(y - k)^2}{b^2} = 1 [/tex]

The Attempt at a Solution

2c = 1, [tex] c = \frac{1}{2} [/tex]

e = c/a, so .5/1 = 1/2 [tex] \frac{(x)^2} {1^2} + \frac{(y)^2}{b^2} = 1 [/tex]

Not quite sure to go from here

 

[SteamKing]

It's not clear what the parameter c is, but the eccentricity can also be defined in terms of a and b, the lengths of the semi-major and semi-minor axes.

 

[LCKurtz]

Use ##a^2=b^2+c^2## for an ellipse.

 

[bnosam]

Use ##a^2=b^2+c^2## for an ellipse.

Since ## 2c = 1 ##, then ##c = .5 ##
Making the variable ##a = 1##

##e = 1/2 = .5 / 1##

## (1)^2 = b^2 + (.5)^2 ##
## 1 - .25 = b^2 ##
## b^2 = 3/4 ##

So this would set the answer to:

## x^2 + \frac{3(y^2)}{4} = 1 ##

Correct?

 

[LCKurtz]

Since ## 2c = 1 ##, then ##c = .5 ##
Making the variable ##a = 1##

##e = 1/2 = .5 / 1##

## (1)^2 = b^2 + (.5)^2 ##
## 1 - .25 = b^2 ##
## b^2 = 3/4 ##

So this would set the answer to:

## x^2 + \frac{3(y^2)}{4} = 1 ##

Correct?

Check your ##\frac{y^2}{b^2}##.

 

[bnosam]

Check your ##\frac{y^2}{b^2}##.

Oops

## x^2 + \frac{4(y^2)}{3} = 1 ##


Thanks a million :)