The curve which passes through (1,2) and whose tangent at any point has a slope that is half of slope of the line joining origin to the point of contact, is -
Let the arbitrary point
be x,y.
Now the slope of the line joining the origin and
the point is
=y−0x−0
=yx.
The equation of the curve
y=f(x).
Now slope of the tangent
=dydx
=y2x ... as per the given condition.
Hence
2dyy=dxx
Integrating both sides we get
2lny=lnx+lnc
Now yx=1=2
Hence
2ln(2)=ln(1)+ln(c)
Or
c=22=4
Hence
2lny=lnx+ln4
Or
ln(y2)=ln4x
Or
y2=4x is the required equation.
This is an equation of parabola.