I have the solution with me, but it looks wrong. Please help me out
The given line is $$x=m(y-10)-\frac 1m$$
Comparing it with $$x-h=m(y-k)+\frac am$$
So $$h=0, ~ k=, ~ a=-1$$
Hence the equation is $$x^2=-4(y-10)$$
But the equation of tangent generally used is $y=mx +\frac am $, which only applies to $y^2=4ax$. How can we use it here?
Why do you think it’s wrong? If you reflect a $Y^2=4aX$ type parabola so that it’s an $X^2=4aY$ type instead, the generic equation of a tangent to it undergoes the same reflection. Your approach to the problem was fine.
Your solution is easily verified: From the equation of the tangents we have $y-10=\frac1m(m x+1)$. Substituting this into your parabola equation, $x^2+\frac4{m^2}(mx+1)=0$. The discriminant of this quadratic equation vanishes, just as required for the line to be a tangent.
solve([m^2*(y-10)-m*x-1, diff(m^2*(y-10)-m*x-1,m)],[x,y]);[[x = -2/m,y = (10*m^2-1)/m^2]]which is your parabola $y=10-x^2/4$. – Jan-Magnus Økland Mar 29 '20 at 18:04