(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove the reflective property of the ellipse.

2. Relevant equations

x^2/a^2 + y^2/b^2 = 1 (equation of ellipse)

tan theta = m2-m1/(1+m1m2)

m = (y1-y0)/(x1-x0)

3. The attempt at a solution

I apologize if this is to simple for this forum, not sure where else to put it.

My idea was to draw the tangent to the ellipse, and lines from the Focii F' and F to the tangent. Then prove that the angles formed by the intersection of F' and the tangent, F and the tangent, are equal. It isn't working out as I had hoped.

Using a tangent to the first quadrant of the ellipse, F' is focus at (-c,0) and F is on the right, (c, 0):

[tex]m_T = -\frac{b^2x_0}{a^2y_0}[/tex]

[tex]m_{F} = \frac{y_0}{x_0-c}[/tex]

[tex]m_{F'} = \frac{y_0}{x_0+c}[/tex]

I got the slope of the tangent by differentiating the equation of the ellipse.

[tex]\tan\theta = \frac{\frac{-b^2x_o}{a^2y_0}-\frac{y_0}{x_0-c}}{1-\frac{b^2x_0y_0}{a^2y_o(x_o-c)}}[/tex]

Which, if my algebra is correct, simplifies to

[tex]\tan\theta = \frac{-(b^2x_0(x_0-c) + a^2y_0)}{a^2y_0(x_0-c) - b^2x_0y_0}[/tex]

A similar process leads to the angle between the tangent and the line from the left focus, F', with result:

[tex]\tan\phi = \frac{-(b^2x_0(x_0+c)+a^2y_0)}{a^2y_0(x_0+c)-b^2x_0y_0}[/tex]

Which don't seem to be equal :grumpy:

Can someone point out where I have gone wrong?

Thank you,

Dorothy

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# Reflective property of ellipse.

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