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Math Help - locus- mid-pt of the chord of a ellipse

  1. #1
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    locus- mid-pt of the chord of a ellipse

    i have problem in (c)
    (4m^2+1)x^2 +8m (1-2m)x+4(4m^2-4m -3)=0
    then i have to solve the parametric equations:

    (x-coordinate of M ) x= -\frac{4m(1-2m)}{4m^2+1}
    and y-1 = m(x-2)
    is there any other method beside put m = \frac {y-1}{x-2} back to x= -\frac{4m(1-2m)}{4m^2+1} ?
    here's the question:
    locus- mid-pt of the chord of a ellipse-out0105.jpeg

    thanks!
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by afeasfaerw23231233 View Post
    i have problem in (c)
    (4m^2+1)x^2 +8m (1-2m)x+4(4m^2-4m -3)=0
    then i have to solve the parametric equations:

    (x-coordinate of M ) x= -\frac{4m(1-2m)}{4m^2+1}
    and y-1 = m(x-2)
    is there any other method beside put m = \frac {y-1}{x-2} back to x= -\frac{4m(1-2m)}{4m^2+1} ?
    here's the question:
    Click image for larger version. 

Name:	out0105.jpeg 
Views:	43 
Size:	51.4 KB 
ID:	6647

    thanks!
    Actually since P and Q both lie on the tangent as well as the ellipse, they should satisfy both the line equation and the ellipse equation.

    So substitute y= m(x-2)+1 in the ellipse equation and expand to get the quadratic (4m^2+1)x^2 +8m (1-2m)x+4(4m^2-4m -3)=0 . The solutions for this must be the points P and Q and thus x_1 and x_2 are the roots
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