Q --> A curve has the equation y # (ax ! 3) ln x, where x p 0 and a is a positive constant. The
normal to the curve at the point where the curve crosses the x-axis is parallel to the line 5y + x = 2.
Find the value of a.
My ans --> y = (ax+3)lnx
On x-axis, y = 0
ax + 3 = 0 ? x is -ve ?no soln
But lnx = 0 ? x =1
dy/dx = alnx + (ax+3).(1/x)
dy/dx (m of tangent) = a + 3
what to do next???
5y+x=2 so y=-(1/5)x+2/5 so gradient =-(1/5)
dy/dx=alnx+a+3/x
y=0 when x=1 since y(1)=(A1+3)LN(1)=0
AND DY/DX AT THIS POINT=a*ln1+1+3/1=a+3=-1/5 so a =-1/5-3=-16/5