Find the equation of a parabola tangent to an exterior circle/arc

Question

I am trying to find the equation of a parabola that is tangent to an exterior circle (this is for designing a bell-shaped nozzle). I know the point (0.055, 0.9) lies on the parabola and also that the circle of radius 0.1625 is centered at (0.1875, 0.25). But I am having trouble getting the point of tangency and then finding the parameters for the parabola.

I tried to match derivates of the functions, but ran into some algebra problems. I started with a general parabola ($y=ax^2+bx+c$) and took the derivate ($y'=2ax+b$). Then I got the equation of a circle centered at (h,k) ( $(x-h)^2+(y-k)^2=r^2$ ) and took the derivate through Wolfram Alpha to get $y'=-\frac{x-h}{y-k}$ but then I got stuck trying to equate the two and find the point of intersection.

Please let me know what I could to try and solve this. Thank you!

Answer 1

Unfortunately you do not have enough information. Let's simplify the problem, and suppose that the parabola is symmetric around the vertical axis. Then the formula for it is $$y=ax^2+c$$ You still have the following unknowns $a$, $c$, $x_i$, $y_i$. The last two are coordinates of the intersection point. But you have only two equations:

• The two curves intersect
• The two curves are tangent

The solution to your problem is unique only if you fix two of those variables. You can fix for example the intersection point. Or you can fix the shape of the parabola.

Answer 2

Making the hypothese that your parabola has equation :

$$y=ax^2+c\tag{1}$$

Taking into account that

$$0.9=a(0.055)^2+c\tag{2}$$

by difference between (1) and (2), we can eliminate $c$, and obtain the following equation for the parabola :

$$y-0.9=a(x^2-0.055^2)\tag{3}$$

Now consider a parametric representation of the current point of the circle :

$$M_{\alpha} = \binom{x(\alpha)=0.1875+0.1625 \cos \alpha}{y(\alpha)=0.25+0.1625 \sin \alpha}\tag{4}$$

The tangent vector at point $M_{\alpha}$ is obtained by differentiation of (4) :

$$T = \binom{x'(\alpha)=-0.1625 \sin \alpha}{y'(\alpha)=0.1625 \cos \alpha}\tag{5}$$

Let us consider that the contact point $M_0$ corresponds to the value $\alpha_0$ of the parameter, i.e.,

$$M_0=(x(\alpha_0),y(\alpha_0))\tag{6}$$

The tangent vector to the parabola at $M_0$ is

$$U = \binom{1}{2a x(\alpha_0)}\tag{7}$$

Let us express that the parabola passes through $M_0$ ; using (3), we have :

$$y(\alpha_0)-0.9=a(x(\alpha_0)^2-0.055^2)\tag{8}$$

from which we can extract

$$a=\dfrac{y(\alpha_0)-0.9}{x(\alpha_0)^2-0.055^2}\tag{9}$$

Besides, expressing that tangent vectors $T$ and $U$ are proportional is equivalent to express the equality of slopes :

$$-\dfrac{\cos \alpha_0}{\sin \alpha_0}=\dfrac{2 a x(\alpha_0)}{1}\tag{10}$$

which can be simplified into

$$a=\dfrac{\text{cotan} \ \alpha_0}{-2 x(\alpha_0)}\tag{11}$$

Relating the two expressions of $a$ in (9) and (11), using relationships (4) end up into the following trigonometric equation :

$$\text{cotan} \ \alpha_0=-2\dfrac{(0.25+0.1625 \sin \alpha_0)-0.9}{(0.1875+0.1625 \cos \alpha_0)^2-0.055^2} (0.1875+0.1625 \cos \alpha_0)\tag{12}$$

that can be solved in different ways.

I have in $(\pi/2,\pi)$ found the unique solution :

$\alpha_0=3.068462$ radians = 175.81 (decimal) degrees

giving in particular

$M_0 (0.0254, 0.2619)$

$a = 268.3$ which is incredibly straight... (almost a vertical line)

Conclusion : as my result has been checked by @Blue, I think that, by modifying the numerical values, but keeping the same final equation (12), you could obtain the nozzle shape with a satisfying curvature.

Remark : considering parabolas with equation $y=ax^2+bx+c$, we have too many parameters...

Answer 3

As noted in @Andrei's answer, the given information is insufficient. @JeanMarie has a detailed answer, so I may not need to post on my own, but I wanted to show that the curviness of the result won't be quite so dramatic as indicated in OP's sketch.

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Below are situations where the parabola meets the circle at a point corresponding to an auxiliary angle relative to the circle's center.

At about $144^\circ$, the parabola is effectively a double-ray pointing straight up; at about $177^\circ$, the parabola degenerates to a straight line tangent to the circle. (Beyond $177^\circ$, the parabola opens downward.)

At about $175.75^\circ$, the parabola is symmetric about the $y$-axis :

Here are a couple more cases: