# What is the equation of ellipse, if given eccentricity=3/4 and distance between two foci is 8 units?

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### 1 Answer

*What is the equation of an ellipse with eccentricity 3/4, and distance between the two foci 8?*

(1) ` `Assume the ellipse is centered at the origin, with major and minor axes along the x and y axes.

(2) Let `f` be the distance from the origin to a focus. So `f=4`

(3) Let `a` be 1/2 the length of the major axis, and `b` 1/2 the length of the minor axis. Let `e` be the eccentricity, `e=3/4`

(4) `e=f/a` (See reference) . So `3/4=4/a => a=(16)/3`

(5) `f=sqrt(a^2-b^2)` (See reference). So `4=sqrt(((16)/3)^2-b^2)`

Then `16=(256)/9-b^2 => b^2=(112)/9=>b=(4sqrt(7))/3` . (Take the positive root as we are finding a length which is positive.)

**(6) The equation of an ellipse is `x^2/a^2+y^2/b^2=1` .**

**`a^2=(256)/9,b^2=(112)/9` so the equation is**

**`x^2/((256)/9)+y^2/((112)/9)=1` or `(9x^2)/(256)+(9y^2)/(112)=1` .**