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Math Help - Find equation of ellpise

  1. #1
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    Find equation of ellpise

    Use the locus definition of an ellipse to find the equation of the ellipse with focus (3,4), corresponding directrix line x+y=1 and eccentricity 1/2.

    so e=1/2, one focus(3,4)
    let the center be (c,d)
    then half the length of the major axis=a
    distance of focus from center=ae distance of directrix from center =a/e.
    center is on the line y=x+1
    half distance between focus and directrix = \frac{1}{2}\frac{3+4-1}{\sqrt{2}}=\frac{3\sqrt2}{2}

    from ae, \frac{1}{2}a=\sqrt{(c-3)^2+(d-4)^2}
    a=2\sqrt{(c-3)^2+(d-4)^2}
    from a/e, 2a=\frac{c+d-1}{\sqrt{2}} finding the length of a perpendicular from a point to a line.
    so, 4\sqrt{(c-3)^2+(d-4)^2}=\frac{c+d-1}{\sqrt{2}}
    i work it out to get 31c^2+31d^2-98c-258d-2cd+799=0
    also a=\frac{c+d-1}{\sqrt{2}}-\frac{3\sqrt2}{2}
    i work that out to be a=\frac{c\sqrt2+d\sqrt2-4\sqrt2}{2}

    Thanks!
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  2. #2
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    One point M on the deretrix is (0,1). From this point draw a perpendicular line to the diretrix. Let this line meet the ellipse at P (x,y). Let S be the focus.
    According to the definition of the ellipse, SP/PM = e
    Find SP and PM and substitute in the above equation and simplify to get the required equation of the ellipse.
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  3. #3
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    Hello arze
    Quote Originally Posted by arze View Post
    Use the locus definition of an ellipse to find the equation of the ellipse with focus (3,4), corresponding directrix line x+y=1 and eccentricity 1/2.
    I think the method you're recommended to use is like this.

    Suppose that P (h,k) is a general point on the ellipse. Then its distance from the focus F(3,4) is:
    FP = \sqrt{(h-3)^2+(k-4)^2}
    and its distance from the line x + y -1 = 0 ( =PD, say) is:
    PD = \frac{|h+k-1|}{\sqrt{1^2+1^2}}= \frac{|h+k-1|}{\sqrt{2}} (See here for this formula.)
    So, using the locus definition of the ellipse:
    \frac{FP}{PD}=e=\frac12

    PD=2FP

     \Rightarrow |h+k-1|=2\sqrt2\sqrt{(h-3)^2+(k-4)^2}

     \Rightarrow (h+k-1)^2=8\Big((h-3)^2+(k-4)^2\Big)

    \Rightarrow h^2+2hk+k^2 - 2h-2k+1=8(h^2-6h+9+k^2-8k+16)

     \Rightarrow 7h^2-2hk+7k^2-46h-62k+199=0
    So the equation of the locus of P is:
      7x^2-2xy+7y^2-46x-62y+199=0
    Grandad
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  4. #4
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    well, this means i got all mixed up! haha thanks very much!
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