If roots of equation (h^2-a^2)x^2-2hkx+(k^2-b^2)=0 are real and equal, prove that h^2/a^2 + k^2/b^2 = 1

We have the equation (h^2-a^2)x^2-2hkx+(k^2-b^2)=0
It is given that the roots of this equation are real and equal, hence D should be equal to 0

D= 0
=> b^2 - 4ac = 0
=> (-2hk)^2 - 4(h^2 - a^2)(k^2 - b^2) = 0
=> 4h^2.k^2 -4(h^2.k^2 - h^2.b^2 - a^2.k^2 + a^2.b^2) = 0
=> 4h^2.k^2 -4h^2.k^2 + 4h^2.b^2 + 4a^2.k^2 - 4a^2.b^2 = 0
=> 4h^2.b^2 + 4a^2.k^2 - 4a^2.b^2 = 0
=> h^2.b^2 + a^2.k^2 - a^2.b^2 = 0

Now, by dividing the equation by a^2.b^2, we'll get

=> h^2.b^2/a^2.b^2 + a^2.k^2/a^2.b^2 - a^2.b^2/a^2.b^2 = 0/a^2.b^2
=> h^2/a^2 + k^2/b^2 - 1 = 0
=> h^2/a^2 + k^2/b^2 = 1