Ellipse

Geometry Level pending

If the locus of the midpoint of the of chords to the ellipse (x/a)^2+(y/b)^2=1 passing through the point (2a,0) is ((x-a)/a)^m+(y/b)^n=1, find m+n


The answer is 4.

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1 solution

Hosam Hajjir
Oct 13, 2017

Let ( x , y ) (x,y) be on the ellipse, then

x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1

The midpoint of the chord between ( 2 a , 0 ) (2a, 0) and ( x , y ) (x, y) is ( x , y ) = ( a + x 2 , y 2 ) (x', y') = ( a + \dfrac{x}{2} , \dfrac{y}{2} ) .

Hence, ( x , y ) = ( 2 ( x a ) , 2 y ) (x, y) = ( 2 (x' - a), 2 y' ) . Substitute this into the equation of the ellipse, you get,

4 ( x a ) 2 a 2 + 4 y 2 b 2 = 1 \dfrac{ 4(x'-a)^2}{a^2} + \dfrac{4 y'^2}{b^2} = 1 . Therefore, m = 2 m = 2 , and n = 2 n = 2 , making the answer 2 + 2 = 4 2 + 2 = 4 .

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