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

Question

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.

Answer 1

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 $$\tan(\pi-2\arctan\sqrt8)=-\frac{2\cdot\sqrt8}{1-(\sqrt8)^2}=\frac{4\sqrt2}{7}.$$

Answer 2

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 $\pi-2\arctan\sqrt8$.