Find the equation of the tangent to the curve `sqrtx+sqrt y = a` at the point `(a^2/4,a^2/4)`

Options

A. `xy=a^(2)`
B. `x+y=(a^(2))/2`
C. `xy=(a^(2))/2`
D. `x-y=(a^(2))/2`

Answer

B

Explanation

Given `sqrt(x)+sqrt(y)=a`
On differentiating both sides w.r.t `x` we get
`1/(2sqrt(x))+1/(2sqrt(y))y'=0`
`implies 1/(2sqrt(x))=-(y')/(2sqrt(y))implies-y/(sqrt(x))=y'`
`:.y'at ((a^(2))/4,(a^(2))/4)=-(sqrt((a^(2))/4))/(sqrt((a^(2))/4))=-1`
Equation of tangent is
`y-(a^(2))/4=-(x-(a^(2))/4)`
`impliesy-(a^(2))/4=-x+(a^(2))/4impliesx+y=(a^(2))/2`