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October 14th, 2015, 11:09 AM   #1
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Parabola Tangent to an Ellipse

I need help finding the 2 points where a parabola is tangent to an ellipse for a project I am working on so that there is a smooth transition from the ellipse to the parabola.

The formula for the ellipse based on my dimensions is x^2/75^2 + y^2/95^2=1

The vertex of the parabola needs to pass through the point (0,-137.5).

Through trial and error I have gotten close, but the parabola either passes through the ellipse at 4 points or none. I know I need to the find where there are just 2 solutions (x,y) and (-x,y).

Any help is greatly appreciated as this is a definitely out of my element.
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October 14th, 2015, 12:22 PM   #2
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$\displaystyle y=(0.0210582)x^2-137.5$


$\displaystyle y=(0.00338627)x^2-137.5$

I think it is the first one.
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October 15th, 2015, 09:27 PM   #3
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Hi Architect,

I think that mrtwhs is right and the correct answer is the first equation given. I'm not sure how much accuracy you need, but I got:

$\displaystyle y=0.02105817571203157 \ x^2-137.5$

and the graphs intersect at ( 68.70567318519605,-38.09552323964478 ) and ( -68.70567318519605,-38.09552323964478 ).

The derivatives (for both the ellipse and the parabola) at these points are 2.8936322766945493 and -2.8936322766945493, respectively.

There was a problem similar to this on the forum a while ago involving finding the equation of the line tangent to a quartic at 2 points if I remember correctly. I did a quick search but could not find it or else I would reference it here. Anyway, I was able to use the same type of analysis to arrive at the answer to your problem:

The equation of the ellipse gives:

$\displaystyle y^2=95^2-\frac{95^2}{75^2}x^2$

and for the parabola:

$\displaystyle y=kx^2-137.5 \quad \Rightarrow \quad y^2=k^2x^4-275kx^2+137.5^2$

where we need to find the constant k. If we subtract the parabola from the ellipse we get:

$\displaystyle 95^2-\frac{95^2}{75^2}x^2-k^2x^4+275kx^2-137.5^2$

The zeroes of this function give the intersection points. Since we are interested in the zeroes, I will divide through by $-k^2$ and rearrange:

$\displaystyle x^4+\left(\frac{95^2}{75^2}-275k \right)\frac{1}{k^2}x^2+\frac{137.5^2-95^2}{k^2}$

Through trial and error I have gotten close, but the parabola either passes through the ellipse at 4 points or none. I know I need to the find where there are just 2 solutions (x,y) and (-x,y).
Yes, and this is the key: At the desired points the function will have two double zeroes, one at $x=B$ and one at $x=-B$.


$\displaystyle x^4+\left(\frac{95^2}{75^2}-275k \right)\frac{1}{k^2}x^2+\frac{137.5^2-95^2}{k^2}=(x-B)^2(x+B)^2=x^4-2B^2x^2+B^4$

From the $B^4$ term we get:

$\displaystyle B^2=\frac{\sqrt{137.5^2-95^2}}{k} \ $ and from the $-2B^2x^2$ term we get:

$\displaystyle B^2=\frac{275}{2k}-\frac{95^2}{2 \cdot 75^2 \cdot k^2}$

Equating the two, multiplying by $k^2$ and solving for $k$ gives:

$\displaystyle k=\frac{95^2}{2 \cdot 75^2} \cdot \frac{1}{137.5-\sqrt{137.5^2-95^2}} \approx 0.021058176$

From the $B^2$ equation given from the $B^4$ term we see that $B$, the $\pm x$ values are given by:

$\displaystyle B=\sqrt{\frac{\sqrt{137.5^2-95^2}}{k}} \approx 68.70567319$

For these $\pm x$ values the y values of the ellipse and the parabola may be shown to be very close to the same.

The derivative of the parabola is $\large \frac{dy}{dx}=2kx$ and the derivative for the ellipse is $\large \frac{dy}{dx}=-\frac{95^2}{75^2}\frac{x}{y}$. The derivative values at the $\pm x$ values may also be shown to be very close the the same, confirming the solution.
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