The equation of parabola is #x^2-4y-2x+5=0#
or #4y=(x^2-2x+1)+4#
or #4(y-1)=(x-1)^2#
and its parametric form #(x(t),y(t))# is #(2t+1,t^2+1)#
Hence, slope of tangent is #((d(t^2+1))/(dt))/((d(2t+1))/(dt))=(2t)/2=t#
and slope of normal is #-1/t#
Therefore equation of normal is
#(y-t^2-1)=-1/t(x-2t-1)#
or #ty-t^3-t=-x+2t+1#
and as it passes through #(h,k)#, we have
#tk-t^3-t=-h+2t+1#
or #t^3+t(3-k)-(h-1)=0#
Let the roots of this be #m_1,m_2# and #m_3# and their product is #h-1#.
Let #m_1# and #m_2# be perpendicuar to each other i.e. #m_1m_2=-1# and hence #m_3=1-h# and as it is root of #t^3+t(2-k+1)-(h-1)=0#, we have
#(1-h)^3+(1-h)(3-k)-(h-1)=0#
and dividing by #1-h# we get
#(1-h)^2+3-k+1=0#
or #2-2h+h^2+3-k=0#
and replacing #(h,k)# by #(x,y)#, we get
#x^2-2x+5-y=0#
or #y=(x-1)^2+4#
graph{(x^2-4y-2x+5)(x^2-2x+5-y)=0 [-9.29, 10.71, -0.08, 9.92]}
