Question

If the slope of the tangent to the curve $y=x\log x$ at a point on it is $\frac{3}{2}$, then find the equations of tangent and normal at that point.

Answer 1

Let the point be $(a,b)$

$y=xlnx$

$b=alna$ ...(1)

also $\frac{dy}{dx|(a,b)}=\frac{3}{2}$

$y=xlnx\frac{dy}{dx}=lnx+1{|}_{x=a}=1+lna$

$1+lna=\frac{3}{2}$

$lna=\frac{1}{2}$

$a=e^{1/2}$ $b=alna$

$=e^{1/2}\frac{1}{2}$

$b=\frac{e^{1/2}}{2}$

equation of tangent

$(y-b)=\frac{dy}{dx}{\int }_{a,b}^{x-a}$

$(y-\frac{e^{1/2}}{2})=\frac{3}{2}(x-e^{1/2})$

$2y-e^{1/2}-3x+3e^{1/2}$

$2y-3x+2e^{1/2}=0$

equation of tangent

$(y-\frac{e^{1/2}}{2})=\frac{-2}{3}(x-e^{1/2})$

$6y-3e^{1/2}=-4x+4e^{1/2}$

$6y+4x=7e^{1/2}$