Results 1 to 2 of 2

Thread: Mid-point of chord on ellipse

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Mid-point of chord on ellipse

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

    I have proven the first part. I got the equation of the tangent at $\displaystyle P_2$ 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 $\displaystyle P_1P_2$, $\displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by arze View Post
    Prove that the equation of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\displaystyle P_1(x_1,y_1)$ is $\displaystyle \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$. The tangent at $\displaystyle P_1$ meets the tangent at $\displaystyle P_2(x_2,y_2)$ at T. Show that the line
    $\displaystyle \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2} +\frac{yy_2}{b^2}$
    passes through T and through the midpoint of $\displaystyle P_1P_2$. Prove that if $\displaystyle P_1TP_2$ is a right angle, then $\displaystyle \frac{x_1x_2}{a^4}+\frac{y_1y_2}{b^4}=0$

    I have proven the first part. I got the equation of the tangent at $\displaystyle P_2$ 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 $\displaystyle P_1P_2$, $\displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.
    The equation $\displaystyle \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2} +\frac{yy_2}{b^2}$ represents a straight line. If the point (x,y) lies on the tangent at $\displaystyle P_1$ then the left side of that equation will be 1. If (x,y) lies on the tangent at $\displaystyle P_2$ 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 $\displaystyle P_1P_2$ to lie on the line is $\displaystyle \frac{\frac12(x_1+x_2)x_1}{a^2}+\frac{\frac12(y_1+ y_2)y_1}{b^2}=\frac{\frac12(x_1+x_2)x_2}{a^2}+\fra c{\frac12(y_1+y_2)y_2}{b^2}$. After multiplying out all the brackets and doing some cancellation, you should be able to write that condition as $\displaystyle \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2} = \frac{x_2^2}{a^2}+\frac{y_2^2}{b^2}$. Both sides of that equation are equal to 1 because $\displaystyle P_1$ and $\displaystyle P_2$ lie on the ellipse. Therefore the condition is satisfied, and the midpoint of $\displaystyle P_1P_2$ lies on the line.

    For the last part of the question, write down the condition for the tangents at $\displaystyle P_1$ and $\displaystyle P_2$ to have perpendicular slopes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find an edge point of ellipse
    Posted in the Math Software Forum
    Replies: 5
    Last Post: Apr 11th 2010, 06:25 AM
  2. how calculate chord length of ellipse.
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Feb 2nd 2010, 02:54 AM
  3. Replies: 3
    Last Post: Feb 27th 2009, 07:52 AM
  4. locus- mid-pt of the chord of a ellipse
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Jun 5th 2008, 05:39 AM
  5. point on an ellipse
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Sep 8th 2006, 07:44 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum