Values of $k$ for which the line $y=kx-1$ is tangent to the parabola with the equation $y=x^{2}+3$

Question

How can I find the values of $k$ for which the straight line $y=kx-1$ is tangent to the parabola with the equation $y=x^{2}+3$?

I used this short cut form $c=-am^{2}$, which gives me $k=\pm 2$. I think I am not correct. What's the answer?

Answer 1

Notice,

Solving the equation of straight line: $y=kx-1$ & equation of the parabola: $y=x^2+3$ $$kx-1=x^2+3\iff x^2-kx+4=0$$ Now, the line will touch the parabola if both real roots of the above quadratic equation are equal, hence we have the determinant $$\Delta=(-k)^2-4(1)(4)=0$$ $$k^2=16\iff |k|= 4$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{k=\pm 4}}$$

Answer 2

HINT:

Find the abscissa $(x)$ of the intersections by equating the values of $y$

Now the two values must be same for tangency