October 14th, 2015, 11:09 AM  #1 
Newbie Joined: Oct 2015 From: Los Angeles Posts: 1 Thanks: 0  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. http://i738.photobucket.com/albums/x...psat0gbijj.jpg 
October 14th, 2015, 12:22 PM  #2 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 
Try $\displaystyle y=(0.0210582)x^2137.5$ or $\displaystyle y=(0.00338627)x^2137.5$ I think it is the first one. 
October 15th, 2015, 09:27 PM  #3  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications 
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: $\displaystyle y=0.02105817571203157 \ x^2137.5$ 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: $\displaystyle y=kx^2137.5 \quad \Rightarrow \quad y^2=k^2x^4275kx^2+137.5^2$ where we need to find the constant k. If we subtract the parabola from the ellipse we get: $\displaystyle 95^2\frac{95^2}{75^2}x^2k^2x^4+275kx^2137.5^2$ 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: $\displaystyle x^4+\left(\frac{95^2}{75^2}275k \right)\frac{1}{k^2}x^2+\frac{137.5^295^2}{k^2}$ Quote:
So: $\displaystyle x^4+\left(\frac{95^2}{75^2}275k \right)\frac{1}{k^2}x^2+\frac{137.5^295^2}{k^2}=(xB)^2(x+B)^2=x^42B^2x^2+B^4$ From the $B^4$ term we get: $\displaystyle B^2=\frac{\sqrt{137.5^295^2}}{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: $\displaystyle k=\frac{95^2}{2 \cdot 75^2} \cdot \frac{1}{137.5\sqrt{137.5^295^2}} \approx 0.021058176$ From the $B^2$ equation given from the $B^4$ term we see that $B$, the $\pm x$ values are given by: $\displaystyle B=\sqrt{\frac{\sqrt{137.5^295^2}}{k}} \approx 68.70567319$ 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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ellipse, parabola, tangent 
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