(x/a) cos m + (y/b) sin m=1 and (x/a) sin m- (y/b) cos m=1 , prove that (x^2/a^2) + (y^2/b^2)=2
The equations are,
(x/a) cos m + (y/b) sin m = 1
(x/a) sin m- (y/b) cos m = 1
Take the square of both equations,
(x/a)^2.cos^2(m) + 2(x/a)(y/b).cos(m).sin(m) + (y/b)^2.sin^2(m) = 1
(x/a)^2.sin^2(m) – 2(x/a)(y/b).sin(m).cos(m) + (y/b)^2.cos^2(m) = 1
Now add together the above two equations, to give
(x/a)^2(cos^2(m) + sin^2(m)) + 0 + (y/b)^2(sin^2(m) + cos^2(m)) = 2
(x/a)^2 + (y/b)^2 = 2