We have the parabola
#f(x,y)=a x^2+b x+c - y = 0#
and a tangent line
#L-> y = y_0+m_0(x-x_0)#
where #m_0 = (5-0)/(5-0) = 1#
Substituting into #f(x,y)# we can determine the intersection points between #f(x,y)=0# and #L#
#f(x,y_0+m_0(x-x_0))=c + b x - m_0 x + a x^2 + m_0 x_0 - y_0=0#
and now solving for #x# we have
#x=(-b + m_0 pm sqrt[(b - m_0)^2 - 4 a (c + m_0 x_0 - y_0)])/(2 a)#
but if #L# is tangent to #f(x,y)=0# then the solution must be real and unique so
#(b - m_0)^2 - 4 a (c + m_0 x_0 - y_0)=0# and solving for #x_0#
#x_0=((b - m_0)^2 + 4 a (y_0-c))/(4 a m_0)# and consequently
#y_0 = (4 a c + m_0^2-b^2)/(4 a)#
Concluding
#L-> y = (4 a c + m_0^2-b^2)/(4 a)+m_0(x-((b - m_0)^2 + 4 a (y_0-c))/(4 a m_0))#
NOTE:
Here #a = 1#, #b =-4# and #c = 0# then
#y_0 = (4 *1*0 + 1^2-(-4)^2)/(4*1) = -15/4#
#x_0=((-4 - 1)^2 + 4*1* (-15/4-0))/(4*1*1) = 5/2#
