Find the equation of the parabola for which the line $m^2(y-10)-mx-1=0$ is a tangent for all real values of $m$

Question

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?

Answer 1

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.