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Math Help - Mid-point of chord on ellipse

  1. #1
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    Mid-point of chord on ellipse

    Prove that the equation of the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at the point P_1(x_1,y_1) is \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1. The tangent at P_1 meets the tangent at P_2(x_2,y_2) at T. Show that the line
    \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2}  +\frac{yy_2}{b^2}
    passes through T and through the midpoint of P_1P_2. Prove that if P_1TP_2 is a right angle, then \frac{x_1x_2}{a^4}+\frac{y_1y_2}{b^4}=0

    I have proven the first part. I got the equation of the tangent at P_2 in the same form as the first. Then I made y the subject of both tangents and found x. did it in the same way to find y.
    I found the the line given passed through T, but could not prove that the line passed through the midpoint of P_1P_2, (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}).
    Thanks!
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  2. #2
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    Quote Originally Posted by arze View Post
    Prove that the equation of the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at the point P_1(x_1,y_1) is \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1. The tangent at P_1 meets the tangent at P_2(x_2,y_2) at T. Show that the line
    \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2}  +\frac{yy_2}{b^2}
    passes through T and through the midpoint of P_1P_2. Prove that if P_1TP_2 is a right angle, then \frac{x_1x_2}{a^4}+\frac{y_1y_2}{b^4}=0

    I have proven the first part. I got the equation of the tangent at P_2 in the same form as the first. Then I made y the subject of both tangents and found x. did it in the same way to find y.
    I found the the line given passed through T, but could not prove that the line passed through the midpoint of P_1P_2, (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}).
    The equation \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2}  +\frac{yy_2}{b^2} represents a straight line. If the point (x,y) lies on the tangent at P_1 then the left side of that equation will be 1. If (x,y) lies on the tangent at P_2 then the right side of the equation will be 1. If (x,y) is the point T then both sides of the equation will be 1, and hence the equation will be satified. Therefore T lies on the line with that equation.

    The condition for the midpoint of P_1P_2 to lie on the line is \frac{\frac12(x_1+x_2)x_1}{a^2}+\frac{\frac12(y_1+  y_2)y_1}{b^2}=\frac{\frac12(x_1+x_2)x_2}{a^2}+\fra  c{\frac12(y_1+y_2)y_2}{b^2}. After multiplying out all the brackets and doing some cancellation, you should be able to write that condition as \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2} = \frac{x_2^2}{a^2}+\frac{y_2^2}{b^2}. Both sides of that equation are equal to 1 because P_1 and P_2 lie on the ellipse. Therefore the condition is satisfied, and the midpoint of P_1P_2 lies on the line.

    For the last part of the question, write down the condition for the tangents at P_1 and P_2 to have perpendicular slopes.
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