Find the angle between two tangents drawn from point $(0, -2)$ to the curve $y=x^2$ 
Find the angle between the two tangents drawn from point $(0, -2)$ to the curve $y=x^2$.

This is my attempt:
Let $P(\alpha, \beta)$ be a point on the curve.
$$\therefore \beta = \alpha^2$$
$$\frac{dy}{dx}\quad \text{at}\quad P(\alpha,\beta) = 2\alpha$$
Equation of tangent at P:
$2\alpha x-y=2\alpha^2-\beta
\Rightarrow2\alpha x-y = 2\alpha^2 -\alpha^2$
$$\Rightarrow2\alpha x -y -\alpha^2 =0$$
A/Q $(0, -2)$ should satify this equation.
$\therefore 2\alpha\times0 - (-2) - \alpha^2 = 0\Rightarrow\alpha^2=2$
$$\therefore\alpha=\pm\sqrt2$$
$$\Rightarrow\beta=2$$
Now putting these values to find slope$(m)=\frac{dy}{dx}=2\times\pm\sqrt2$
$$\therefore m = +2\sqrt2\quad and\quad -2\sqrt2$$
We know that for $\theta$=angle between the lines and $m_1\quad\&\quad m_2$ be slope of lines:
$$\tan{\theta} =\big|\frac{m_1-m_2}{1+m_1m_2}\big|$$
$$=\big|\frac{2\sqrt2-(-2\sqrt2)}{1+2\sqrt2\times-2\sqrt2}\big|= \big|\frac{4\sqrt2}{1-8}\big|=\frac{4\sqrt2}7$$
My answer does not match the book. The book is a very appreciated one, so it can't be wrong. I cannot find an error in my solution.
The book states the answer as
$$\pi - 2\arctan\sqrt8$$
Edit: The book actually states its answer as $\pi - 2\arctan\sqrt8$. I was the blind who couldn't see the 2.
 A: I think there are two mistakes in your book.
Firstly, $\theta$ should be an acute angle because we say about angle between tangents and not say about segments of tangents, but $\pi-\arctan\sqrt8>\frac{\pi}{2}.$
Also, your answer $\arctan\frac{4\sqrt2}{7}$ is true and even $\arctan\frac{4\sqrt2}{7}\neq\arctan\sqrt8$.
After your fixing we need to prove that
$$\tan(\pi-2\arctan\sqrt8)=\frac{4\sqrt2}{7},$$ which is true because
$$\tan(\pi-2\arctan\sqrt8)=-\frac{2\cdot\sqrt8}{1-(\sqrt8)^2}=\frac{4\sqrt2}{7}.$$
A: As you have already derived, the $x$-values of the two points on the curve are $-\sqrt2$ and $\sqrt2$ (with a respective $y$-value of $2$).
Let's have a look on the "right tangent" $\big($at $(+\sqrt2,2)$$\big)$. As the slope of the tangent is $2\sqrt2=\sqrt8$, the angle between the $x$-axis and this tangent is $\arctan\sqrt8$. Therefore, the angle between that tangent and the $y$-axis is ${\pi\over2}-\arctan\sqrt8$. Finally, the double of this angle is the angle between the two tangents, which indeed is $\pi-2\arctan\sqrt8$.
