Point P(a,b) is on the curve
square root of x + square root of y =1.
Show that the slope of the tangent at P is : - square root of (b/a).
To find the slope of the tangent at point P(a,b) on the curve, we need to find the derivative of the curve equation with respect to x, and then substitute the values of a and b into the derivative.
The given equation is:
√x + √y = 1
To find the derivative, we need to isolate y on one side of the equation:
√y = 1 - √x
Now square both sides of the equation to eliminate the square root:
y = (1 - √x)^2
y = 1 - 2√x + x
Now we can find the derivative of y with respect to x:
dy/dx = -2(1/2)√x + 1
dy/dx = -√x + 1
Now substitute a and b into the derivative:
Slope = dy/dx = -√a + 1
Therefore, the slope of the tangent at point P(a,b) is -√a + 1.
(Note: In the question, it is mentioned that the slope should be -√(b/a), but according to the given equation, the slope is -√a + 1.)