The function #y=xln(1/x^2)# can be simplified to #y=xln(x^-2)=-2xln(x)#.
A normal line is perpendicular to the tangent line. Thus, we need to find the slope of the tangent line of the function at #x=2# (call it #m#). The slope of the normal line of the function at #x=2# then is given by #-1/m#. We also know that it crosses #-2xln(x)# at #x=2rArry=-2*2ln(2)=-4ln(2)#. Since we know its slope and a point, we can find the equation of the line using the point-slope form #y-y_1=m(x-x_1)#, where #m# is the slope and #(x_1,y_1)# is a point on the slope.
The slope of the function #y=-2xln(x)# at #x=2# can be calculated by using calculus. #dy/dx=-2ln(x)-2# (using the product rule #(d(uv))/dx=v(du)/dx+u(dv)/dx#). The slope of the line, #dy/dx#, at #x=2# is #-2ln(2)-2#. Therefore, the slope of the normal line is #-1/(-2ln(2)-2)=1/(2ln(2)+2)#.
Therefore, the normal line is given by #y+4ln(2)=(x-2)/(2ln(2)+2)#. Simplify this to get #y=x/(2ln(2)+2)-1/(ln(2)+1)-4ln(2)#.
Verify this using the graph below:
graph{(y-xln(1/x^2))(y-x/(2ln(2)+2)+1/(ln(2)+1)+4ln(2))=0 [-10, 10, -5, 5]}
