# Equation of an ellipse given c and eccentricity.

• bnosam
In summary, to write the equation of an ellipse with a center at (0, 0), a horizontal major axis, and eccentricity of 1/2, the equation would be x^2 + 4y^2/3 = 1.
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

$$\frac{(x - h)^2} {a^2} + \frac{(y - k)^2}{b^2} = 1$$

## The Attempt at a Solution

2c = 1, $$c = \frac{1}{2}$$

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

Not quite sure to go from here

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.

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

LCKurtz said:
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?

bnosam said:
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 said:

Oops

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

Thanks a million :)

## 1. What is the equation of an ellipse given c and eccentricity?

The equation of an ellipse given c and eccentricity is x^2/a^2 + y^2/b^2 = 1, where a = c/e and b = a * sqrt(1-e^2).

## 2. How do you calculate the major and minor axes of an ellipse given c and eccentricity?

The major and minor axes of an ellipse can be calculated using the equation a = c/e and b = a * sqrt(1-e^2), where c is the distance from the center to one of the foci and e is the eccentricity of the ellipse.

## 3. Can the eccentricity of an ellipse be greater than 1?

No, the eccentricity of an ellipse can only range from 0 to 1. An eccentricity of 0 represents a circle, while an eccentricity of 1 represents a parabola. Any value greater than 1 would result in a hyperbola.

## 4. How is the eccentricity of an ellipse related to its shape?

The eccentricity of an ellipse determines its shape. A lower eccentricity results in a more circular shape, while a higher eccentricity results in a more elongated shape. The closer the eccentricity is to 1, the more elongated the ellipse becomes.

## 5. Can you graph an ellipse given c and eccentricity?

Yes, you can graph an ellipse given c and eccentricity using the equation x^2/a^2 + y^2/b^2 = 1, where a = c/e and b = a * sqrt(1-e^2). Simply plot the center point at (0,0) and then plot the vertices at (+/- a, 0) and (0, +/- b). Connect these points with a smooth curve to complete the ellipse.

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