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

Explanation

Given curve `sqrt(x)+sqrt(y)=a.....(1) `
we have to seek out the equation of the tangent to the curve at the point `((a^2)/4,(a^2)/4)`
differentiate equation (1) w.r.t `x` we get,
`=>1/(2sqrt(x))+1/(2sqrt(y))(dy)/(dx)=0`
`=>(dy)/(dx)=(-sqrt(y))/sqrt(x)`
Given `(x_1,y_1)=((a^2)/4,(a^2)/4)`
Slop of tangent , `m=((dy)/(dx))_((a^2)/4,(a^2)/4)`
`=(-sqrt((a^2)/4))/sqrt((a^2)/4)`
`=-1`
Equation of tangent,
`y-y_1=m(x-x_1)`
`=>y-(a^2)/4=-1(x-(a^2)/4)`
`=>y-(a^2)/4=-x+(a^2)/4`
`=>x+y=(a^2)/4`
that is the equation of tangent.