# Thread: Finding the cartesian equation of an ellipse

1. ## Finding the cartesian equation of an ellipse

The foci of an ellipse are at S(4,0) and T(-4,0) and for any point P on the ellipse SP+TP=10. Find the cartesian equation of the ellipse.

I tried to tackle the question by giving point P the general coordinates (acosx,asinx) and then working out the lengths SP and TP, equating them to 10, but this gave a really complicated equation which was nothing like the answer: [(x^2)/25]+[(y^2)/9]=1.

2. Focus point is at $\sqrt{a^2-b^2}$ where the following equation of an ellipse holds: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

3. Now I'm confused lol. I thought the foci are at (+/- ae,0)?

4. Originally Posted by free_to_fly
The foci of an ellipse are at S(4,0) and T(-4,0) and for any point P on the ellipse SP+TP=10. Find the cartesian equation of the ellipse.

I tried to tackle the question by giving point P the general coordinates (acosx,asinx) and then working out the lengths SP and TP, equating them to 10, but this gave a really complicated equation which was nothing like the answer: [(x^2)/25]+[(y^2)/9]=1.

Hello,

take the definition of an ellipse as the locus of all points which have the constant sum of distances to 2 fixed points (the focii) of 2a (a is the length of the major semiaxis).

Therefore you know

$2a = 10~\iff~a = 5$

$e = 4~\wedge~e= \sqrt{a^2-b^2}~\implies~ 4=\sqrt{25-b^2} ~\implies~ b=3$

Thus the equation of your ellipse is:

$\frac{x^2}{25} + \frac{y^2}{9}=1$