Prove that the straight lines joining the origin to the point of intersection of #kx+hy=2hk# and the curve #(x-h)^2+(y-k)^2=a^2# are perpendicular if #h^2+k^2=a^2# ?

1 Answer
P dilip_k
Aug 11, 2018

Given equation of the straight line

#kx+hy=2hk#

#or,x/(2h)+y/(2k)=1.....(1)#

Given equation of a curve

#(x-h)^2+(y-k)^2=a^2#

#or,x^2+y^2-2hx-2ky+h^2+k^2-a^2=0.....(2)#

The coordinates of the points in which the straight line meets the curve will satisfy both equations (1) and (2) and hence they satisfy the following equation.

#x^2+y^2-2(hx+ky)(x/(2h)+y/(2k))+(h^2+k^2-a^2)(x/(2h)+y/(2k))^2 =0#

#=>4h^2k^2(x^2+y^2)-4hk(hx+ky)(kx+hy)+(h^2+k^2-a^2)(hx+ky)^2 =0)#

#=>4h^2k^2(x^2+y^2)-4hk(hkx^2+hky^2+(h^2+k^2)xy)+(h^2+k^2-a^2)(kx+hy)^2 =0.#

#=>4h^2k^2(x^2+y^2)-4h^2k^2(x^2+y^2)-4hk(h^2+k^2)xy+(h^2+k^2-a^2)(k^2x^2+h^2y^2)+(h^2+k^2-a^2)(2hkxy) =0.#

#=>cancel(4h^2k^2(x^2+y^2))-cancel(4h^2k^2(x^2+y^2))-4hk(h^2+k^2)xy+(h^2+k^2-a^2)(k^2x^2+h^2y^2)+(h^2+k^2-a^2)(2hkxy) =0.#

This is the equation of the pair of straight lines joining the origin to the points of intersection of the given lines and the curve. They will be at right angle if the sum o bef the coefficients of #x^2andy^2# is zero.

Imposing this condition we get

#(h^2+k^2-a^2)(h^2+k^2) =0#

As #(h^2+k^2) !=0#

We get

#h^2+k^2-a^2=0#

#=>h^2+k^2=a^2#

Related questions
Impact of this question
8019 views around the world
You can reuse this answer
Creative Commons License