differential equation for `y^2=4a(x-a)`

Options

A. `yy'-2xyy' +y^(2)=0`
B. `yy'(yy'+2x) +y^(2)=0`
C. `yy'(yy' -2x) +y^(2)=0`
D. `yy' -2xyy +y =0`

Answer

C

Explanation

Given curve is ` y^(2) = 4a (x-a)`
On differentiating w.r.t,x, we get
2yy' 4a
On putting the value of a Eq, (i) we get
` y^(2) = 4 ((yy')/2) ( x - (yy)/2) = yy' (2x- yy')`
` Rightarrow yy' (yy' -2x) + y^(2) =0`