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I'm a little confused about a problem that asks me to find the angle between the two parabolas $$y^2=2px-p^2$$ and $$y^2=p^2-2px$$ at their intersection. I used implicit differentiation to find the slopes $$y'=\frac{-p}{y}$$ and $$y'=\frac{p}{y}$$ Using the formula for the angle between two lines $\tan\alpha=|\frac{m_2-m_1}{1+m_2m_1}|$ with some substitutions I ended up with $\tan\alpha=|\frac{y}{x}|$. When I set the two equations equal to one another to find the point of intersection I ended up with $x=0$ and $y=±p$.

I know these values lead to tangent being undefined, can I infer from this fact that the angle between these parabolas is a right angle? The solution says that they intersect at right angles, but I'm confused about using something that is undefined to formulate my answer, if it turns out to be correct. Thanks for any help.

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  • $\begingroup$ I'm using the slopes of the lines tangent to the point of intersection for both parabolas. The formula is valid for the slopes of these lines. $\endgroup$ – sylvester Aug 4 '15 at 4:02
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    $\begingroup$ Your two parabolas meet at the common tangent point of their vertices, with the mutual tangent line being "vertical", $ \ x \ = \ \frac{p}{2} \ $ . This may explain your "undefined" tangent value result. $\endgroup$ – colormegone Aug 4 '15 at 4:21
  • $\begingroup$ When set both eqautions equal to one another you will be ending up with $x=p/2,y=0$ $\endgroup$ – QuantumGirl Aug 4 '15 at 4:25
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    $\begingroup$ The point of intersection is at $(p/2,0)$. Inasmuch as $y=0$, one cannot divide by $y$. If one attempts to take a derivative, one obtains $yy'=\pm p$ and thus we see that $y'$ is undefined and the tangent lines are both vertical. Thus the angle between them is zero. $\endgroup$ – Mark Viola Aug 4 '15 at 4:30
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First, let's find the point of intersection. This occurs when

$$p^2-2px_0=2px_0-p^2\implies x_0=p/2\implies y=0$$

At the point of intersection, $y=0$ and thus $yy'=\pm p$ implies that $y'$ is undefined and therefore the tangent to both parabolas is a vertical line in the $x-y$ plane defined be the equation $x=p/2$.

Inasmuch as the tangent lines are parallel, the angle between them is $0$ and we are done!

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Notice, we have $$y^2=2px -p^2$$$$\implies 2y\frac{dy}{dx}=2p \iff \frac{dy}{dx}=\frac{p}{y}=m_1$$

$$y^2=p^2-2px$$$$\implies 2y\frac{dy}{dx}=-2p \iff \frac{dy}{dx}=\frac{-p}{y}=m_2$$ Now, the angle between the parabolas is given as $$\tan \alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ $$=\left|\frac{\frac{p}{y}-\left(\frac{-p}{y}\right)}{1+\frac{p}{y}\left(\frac{-p}{y}\right)}\right|$$ $$=\left|\frac{2py}{y^2-p^2}\right|$$

On solving the equations of the parabolas we get intersection point $\left(\frac{p}{2}, 0\right)$ which is lying on the x-axis.

Now, substituting the value of $y$ we get the angle between the curve at the point of intersection $$\tan \alpha=\left|\frac{2p(0)}{(0)^2-p^2}\right|$$ $$=0 \iff \alpha=\tan^{-1}(0)=0$$ Thus, the angle between the given parabolas at the point of intersection $\left(\frac{p}{2}, 0\right)$ is $\alpha=0$ $$\bbox[5px, border: 2px solid #C0A000]{\color{red}{\alpha=0}}$$ In other words, we can also say that the given parabolas: $y^2=2px-p^2$ & $y^2=p^2-2px$ are touching each other at the point $\left(\frac{p}{2}, 0\right)$.

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    $\begingroup$ You are explicitly dividing by zero all over the solution. $\endgroup$ – Mark Viola Aug 4 '15 at 4:28
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    $\begingroup$ I suspect that the problem was posed with just this sort of treachery intended. $\endgroup$ – colormegone Aug 4 '15 at 4:42
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Ok, ok, let's don't divide by zero. Parametric equations of this curves are $$ C1: \left(\frac{p^2 + y^2}{2p},\, y\right)\\ C2: \left(\frac{p^2 - y^2}{2p},\, y\right) $$ ($p\ne 0$, ok, right? :) Tangent vectors are $$ C1: \left(\frac{y}{p},\, 1\right)\\ C2: \left(-\frac{y}{p},\, 1\right) $$ At point of intersection ($y=0$) both are equal to $(0,\, 1)$, and angle is zero.

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