the answer is √x + √y = √c
=> 1/(2√x) + 1/(2√y) dy/dx = 0
=> dy/dx = - √y/√x
Let (x', y') be any point on the curve
=> equation of the tangent at that point is
y - y' = - (√y'/√x') (x - x')
x-intercept of this tangent is obtained by plugging y = 0
=> 0 - y' = - (√y'/√x') (x - x')
=> x = √(x'y') + x'
y-intercept of the tangent is obtained by plugging x = 0
=> y - y' = - (√y'/√x') (0 - x')
=> y = y' + √(x'y')
Sum of the x and y intercepts
= √(x'y') + x' + y' + √(x'y')
= (√x' + √y')^2
= (√c)^2 (because (x', y') is on the curve => √x' + √y' = √c)
= c. hope this helps :D
