The equation of the tangent to curve √xa+√yb=2 at the point (a, b) is
We have,
√xa+√yb=2.......(1)
√x√a+√y√b=2
On differentiation and we get,
12√x√a+12√y√bdydx=0
12√y√bdydx=−12√x√a
dydx=−√y√b√x√a
At the point (a,b) and we get,
dydx=−√b√b√a√a
dydx=−ba
Equation of tangent is
y−y1=dydx(x−x1)
y−b=−ba(x−a)
ay−ab=−bx+ab
ay+bx=2ab
ay+bxab=2
yb+xa=2
xa+yb=2
Hence, this is the answer.