MHB Finding Distinct and Singular Normals for a Parabola

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The discussion revolves around finding the distinct and singular normals for the parabola defined by the equation y = 2x^2 + 4x + 7/4. Participants are tasked with identifying regions in the xy-plane where there are either three distinct normals or only one normal to the curve. The problem emphasizes the geometric properties of the parabola and the conditions under which normals can be drawn from a given point (X, Y). The challenge is framed as a mathematical puzzle, inviting engagement and problem-solving. The conversation highlights the complexity and interest in the behavior of normals to parabolic curves.
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Let $$y = 2x^2 + 4x + \tfrac{7}{4}$$ and line $$p$$ a normal which go through the point $$(X, Y)$$. Find the regions in $$xy$$-plane where there are either 3 distinct normals or only 1 normal.
 
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Theia said:
Let $$y = 2x^2 + 4x + \tfrac{7}{4}$$ and line $$p$$ a normal which go through the point $$(X, Y)$$. Find the regions in $$xy$$-plane where there are either 3 distinct normals or only 1 normal.

Hey Theia! Nice puzzle! ;)

Let's pick a point $(x,y)$ on the parabola.
Then the vector $\mathbf d$ from $(X,Y)$ to $(x,y)$ is given by:
\[\mathbf d = \binom xy - \binom XY = \binom {x-X}{2x^2+4x+7/4-Y}\]
A tangential vector $\mathbf t$ at $(x,y)$ is the derivative:
\[\mathbf t = \d{}x \binom x{2x^2+4x+7/4}= \binom 1 {4x+4}\]
The corresponding line $p$ along $\mathbf d$ is normal if the dot product is zero:
\[\mathbf d \cdot \mathbf t = (x-X)+(2x^2+4x+7/4-Y)(4x+4) = 8x^3+24x^2+(24-4Y)x+(7-4Y-X) = 0\]
This is a cubic function, and according to wiki, the critical distinction is when its discriminant $\Delta=0$, that is:
\begin{array}{}\Delta &=& 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 \\
&= &18\cdot 8\cdot 24(24-4Y)(7-4Y-X)-4\cdot 24^3(7-4Y-X)+24^2(24-4Y)^2 \\
&& - 4\cdot 8 (24-4Y)^3-27\cdot 8^2(7-4Y-X)^2 \\
&=& -64(27X^2+54X-32Y^3+27) \\
&=& 0\end{array}

If we plot this with Wolfram, we get:
View attachment 5933

In other words, we have 3 distinct normals if we are above that curve.
And we have 1 distinct normal if we are below that curve.

Oh, and the intersection points are at $\left(-1\pm\frac 1{\sqrt 2}, \frac 34\right)$, and the inverting point is at $(-1,0)$.
 

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Very well done! ^^
 
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