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Thread: Ellipse proofs

  1. #1
    Senior Member
    Jul 2009

    Ellipse proofs

    The foci of an ellipse are S, S' and P is any point on the curve. The normal at P meets SS' at G. Prove the $\displaystyle PG^2+(1-e^2)PS.PS'$
    where e is the eccentricity of the ellipse.
    I let P(s,t)
    the equation of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
    where $\displaystyle b^2=a^2(1-e^2) \rightarrow(1-e^2)=\frac{b^2}{a^2}$
    S(ae,0) and s'(-ae,0)
    gradient of normal at P=$\displaystyle -\frac{2a^2y}{b^2(a^2-2x)}$
    equation PG $\displaystyle y-t=-\frac{2a^2t}{b^2(a^2-2s)}(x-s)$
    when y=0 $\displaystyle x=s+\frac{b^2}{a^2}\frac{a^2-2s}{2}$
    $\displaystyle PG^2=(s-(s+\frac{b^2}{a^2}\frac{a^2-2s}{2}))^2+t^2$
    $\displaystyle =(\frac{b^2}{a^2})^2(\frac{(a^2-2s)^2}{4})+t^2$
    $\displaystyle =(1-e^2)^2(\frac{(a^2-2s)^2}{4})+t^2$

    PS.PS'=$\displaystyle \sqrt{(s-ae)^2+t^2}\times\sqrt{(s+ae)^2+t^2}$
    Now my problem is here, i cannot find a square root for this expression.
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  2. #2
    Super Member
    Jun 2009
    First of all let us see what will be PS*PS'.
    PS = e*PM, where PM is the distance of P from the diretrix.
    PS' = e*PM'. If co-ordinates of P are (x1, y1),
    PM = (a/e - x1) and PM' = (a/e + x1). Hence
    PS*PS' = e^2(a^2/e^2 - x1^2) = (a^2 - e^2*x1^2)........(1)
    The equation of the tangent to the ellipse is given by
    xx1/a^2 + yy1/b^2 = 1
    Slope of the tangent is -x1*b^2/y1*a^2
    Slope of the normal is y1a^2/x1b^2
    Find the equation of the normal.
    Find the point of intersection of the normal with x-axis.
    Find PG^2.
    Replace y1^2 in the above expression by b^2(1 - x^2/a^2)
    Now simplify.
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