Originally Posted by

**Grandad** Although you haven't said so, I'm guessing that the lines $\displaystyle y=\pm 2x - 2$ are tangents to the parabola. So, on that assumption, we can find the tangent at the point $\displaystyle (x_1, kx_1^2)$, and then compare it to one of these equations. Like this:

$\displaystyle y = kx^2$

$\displaystyle \Rightarrow \frac{dy}{dx}= 2kx =2kx_1$ at $\displaystyle (x_1, kx_1^2)$

So the tangent at this point is

$\displaystyle y-kx_1^2 =2kx_1(x-x_1)$

i.e. $\displaystyle y =2kx_1x -kx_1^2$ This is the equation of the tangent to the parabola at any general point $\displaystyle \color{red}(x_1, kx_1^2)$ on the parabola. Do you understand how I got this?

So if this is the same as the line $\displaystyle y = 2x -2$ (i.e. one of the two lines you were given in the question), we can compare coefficients to get:

Coefficients of $\displaystyle \color{red}x$ in the two equations are the same: $\displaystyle 2kx_1 =2$ and the constant terms are the same: $\displaystyle -kx_1^2 = -2$

$\displaystyle \Rightarrow kx_1= 1$ and $\displaystyle kx_1^2 =2$

Divide the second equation by the first: $\displaystyle \color{red}\frac{kx_1^2}{kx_1} = \frac{2}{1}$

$\displaystyle \Rightarrow k =0.5$ and $\displaystyle x_1 = 2$

Can you complete it now?

Grandad