Use the discriminant to show that $mx−y + m^2 = 0$ touches the parabola $x^2 =−4y$, for all values of m.

Question

Use the discriminant to show that $mx−y + m^2 = 0$ touches the parabola $x^2 =−4y$, for all values of m.

I attempted to solve by letting them both equal each other, but it didn't work. How do I do this question?

Thank You in advance.

Answer 1

Given the equation of the straight line $$y=mx+m^2$$ & parabola $$x^2=-4y$$ solving both the equations for intersection points, we get $$x^2=-4(mx+m^2)$$ $$x^2+4mx+4m^2=0$$ Now, to find the nature of roots of above quadratic equation, lets' check the determinant as follows $$\Delta=B^2-4AC=(4m)^2-4(1)(4m^2)=16m^2-16m^2=0$$ Since $\Delta=0$ then both roots of above quadratic equation will be equal.

Hence the given line intersects the parabola at a single point i.e. the given line: $y=mx+m^2$ touches the parabola: $x^2=-4y$ for all real values of $m$

Answer 2

So the two functions are $y=-\frac{1}{4}x^2$ and $y=mx+m^2$.

Setting them equal, we have $-\frac{1}{4}x^2=mx+m^2$

which rearranges to $0=x^2+4mx+4m^2$

then you can use the quadratic formula (or simply complete the square) to get the result. I suspect that this is where you are supposed to use the discriminant, since in the case the discriminant would be $16m^2-16m^2=0$, showing that there is always one solution.