Show that $\forall k, y = \frac{x}{k} + \frac{k}{4} $ is a tangent to $y^2 = x$

Question

Can someone please give an intuition on how to start? I was thinking of differentiating the $y^2$ term but I’ve no idea what to do after that.

Answer 1

$y^2=x$ gives $\dfrac{dy}{dx} = \dfrac{1}{2y}$.

Take an arbitrary point $(a^2,a)$ on the curve. The tangent line there is given by

$$y -a = \dfrac{1}{2a}(x-a^2)$$

or

$$y = \dfrac{1}{2a}x+\dfrac{a}{2}$$

You can see that choosing $a=\dfrac{k}{2}$ gives the desired result for any $k$.

Answer 2

We must assume $k\neq0$ unless we rewrite the line’s equation to $4ky=4x+k^2$. If $k=0$, the line and the parabola meet only in $(0,0)$.

Let $k\neq0$. Now substitute $y=x/k+4/k$ in the second equation. The only solution is $x=k^2$ and $y=k+4/k$. Since there is only one solution the straight line is tangent to the parabola.