Limiting Angle

There must be other ways to solve this, but I found this method quite traditional.

First of all,
Letting $\angle APO = \alpha$ ,
If $\displaystyle\sec\alpha$ has a limit, then $\displaystyle\sin\alpha$ also has a limit. Thus, we shall first find \displaystyle\sin\alpha).

Consider a general point $\displaystyle P(x,y)$ , which lies on the given parabola.
It is clear that,
$\displaystyle y = 2x(x-1)$

Using Sine Rule,
$\displaystyle\frac{\sin\alpha}{OA} = \frac{\sin\angle AOP}{AP}$

Thus,
$\displaystyle\frac{\sin\alpha}{1} = \frac{\frac{y}{\sqrt{x^2+y^2}}}{AP} = \frac{\frac{y}{\sqrt{x^2+y^2}}}{\sqrt{(x-1)^2+y^2}}$

Substituting $\displaystyle y=2x(x-1)$ , we get,
$\sin\alpha=\frac{2x(x-1)}{\sqrt{(4x^4-8x^3+5x^2)(4x^4-8x^3+5x^2-2x+1)}}=\frac{2(x-1)}{\sqrt{(4x^2-8x+5)(4x^4-8x^3+5x^2-2x+1)}}$

Note that we cancelled out the $\displaystyle x$ 's, which is why we could not just substitute the final x-coordinates i.e $\displaystyle x=0$

Now, substituting $\displaystyle x=0$ , because, as the problem suggests, finally, $\displaystyle x\to 0$

$\displaystyle\sin\alpha = \frac{-2}{\sqrt{5}}$

$\displaystyle\Rightarrow\cos\alpha = \pm\frac{1}{\sqrt{5}}$

We reject the positive value, because, when $\displaystyle P$ approaches $\displaystyle O$ , clearly, $\displaystyle\angle APO > 90^o$
$\displaystyle\Rightarrow\sec(APO)<0$

Thus,
$\displaystyle\sec\alpha = -\sqrt{5}$

$\text{Recall: } \\ \text{The slope } m(\overrightarrow{PO}) \text{ of the secant line } \overrightarrow{PO} \text{ is the average rate of change between } P \text{ and } O \\ \lim_{_{P \to O}} m(\overrightarrow{PO}) = \text{Slope of the tangent line at point } O \\$

$\text{What the problem is essentially asking is the secant of the angle between } \\ \text{the tangent line at point O and the line } \overrightarrow{OA}.$

http://postimg.org/image/ssy8jb1ll/

$\mbox{ } \\ \text{To solve the problem we need to find the angle } \theta \text{ using the dot product of } \mathbf{T} \text { and } \mathbf{OA}. \\ \text{To do this we must first find vector } \mathbf{T}.$

$m(x) = \frac{\mathrm d}{\mathrm d x} \big( 2x(x -1) \big) = 4x - 2 \quad \text{and} \quad m(0) = -2 \\ \mathbf{T} = \langle -1, 2 \rangle \quad \text{and} \quad \mathbf{OA} = \langle 1, 0 \rangle \\ _{\text{Note: To find vector T just choose two points on the line y = -2x, e.g (-1,2) and (0,0), and subtract their components.}} \\$

$\text{Now that we have the two vectors we can solve the problem. }$

$\mathbf{T}\bullet \mathbf{OA} = -1 \quad \quad | \mathbf{T} | = \sqrt{5} \quad \quad | \mathbf{OA} | = 1 \\ \mathbf{T}\bullet \mathbf{OA} = | \mathbf{T} | \mbox{ } | \mathbf{OA } | \mbox{ } cos(\theta ) \quad \rightarrow \quad cos(\theta) = {{\mathbf{T} \bullet \mathbf{OA}} \over {| \mathbf{T} | | \mathbf{OA}|}} = {-1 \over \sqrt{5}} \\$

$\text{Finally:} \\ sec(\theta) = {1 \over cos(\theta)} = \boxed{- \sqrt{5}}$

My solution involves finding $\angle APO$ in terms of the co-ordinates of point $P$ and then taking the limit as $P \to O$ .

Let $P \equiv (t, 2t(t-1))$ , as it lies on the parabola $y=2x(x-1)$ .

Slope of $OP, m_1 = \frac{2t(t-1)-0}{t-0} = 2(t-1)$ .

Slope of $PA, m_2 = \frac{2t(t-1)-0}{t-1} = 2t$ .

Let $\angle APO = \theta$ , then $\left|\tan \theta \right|= \left| \frac{m_2-m_1}{1+m_1m_2}\right|$ .

Since $\sec^2 \theta = 1 + \tan^2 \theta$ , we have $\sec \theta = \pm \sqrt{1 + \left(\frac{2}{1+4t(t-1)}\right)^2}$ .

Observe that as $P \to O, t \to 0$ ; and also $\angle APO > 90^\circ$ , so $\sec \angle APO < 0$ .

$\lim_{P \to O} \sec \angle APO = - \lim_{t \to 0} \sqrt{1 + \left(\frac{2}{1+4t(t-1)}\right)^2} = -\sqrt{5}$ .