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Math Help - Finding the cartesian equation of an ellipse

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    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.

    Can someone please help me with this problem? Thanks in advance.
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  2. #2
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    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
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    Now I'm confused lol. I thought the foci are at (+/- ae,0)?
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    Quote Originally Posted by free_to_fly View Post
    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.

    Can someone please help me with this problem? Thanks in advance.
    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
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