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 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 $\displaystyle y=(0.0210582)x^2-137.5$ or $\displaystyle y=(0.00338627)x^2-137.5$ I think it is the first one. 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:

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

Quote:
 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 $x=-B$.

So:

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

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. Tags ellipse, parabola, tangent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post elim Geometry 5 September 19th, 2014 02:11 PM lauchagonzalez Calculus 9 March 23rd, 2014 12:44 AM Gluon Trigonometry 6 March 18th, 2014 11:21 AM rollerJ Real Analysis 0 February 23rd, 2012 02:07 AM smash Algebra 5 December 30th, 2010 09:08 PM

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