Question

If the tangent is drawn to the curve y = f(x) at a point where it crosses the y - axis then its equation is

• A. $x-4y=2$
• B. $x+4y=2$
• C. $x+4y+2=0$
• D. none of these

Answer 1

The correct option is A $x-4y=2$

It is given that
$\frac{df\left(x\right)}{dx}=f(x{)}^2$
Let
$f\left(x\right)=y$
Then
$y^{\prime }=y^2$
Or
$\int y^{-2}.dy=\int dx$
Or
$\frac{-1}{y}=x+c$
Now
$y=\frac{-1}{2}$ at x=0.
Hence
$c=2$
Or
$-\frac{1}{y}=x+2$
Or
$xy+2y+1=0$ ...(i)
${\frac{dy}{dx}}_{x=0}$
$=y_{y=\frac{-1}{2}}^2$
Or
$y^{\prime }=\frac{1}{4}$.
Now equation of the tangent is
$y-\left(\frac{-1}{2}\right)=\frac{1}{4}(x-0)$
Or
$4y+2=x$
Or
$x-4y=2$ is the required equation of tangent.