Ellipse vs Parabola: Find a, b & c at x=±4

[roam]

Lets consider an ellipse with equation \frac{x^2}{25}+y^2=1 and a parabola with equation y=ax²+bx+c (on the same grid) meet at x = ±4 (y>0).

This is in a way that at both points would have identical tangents, of course.

In this situation, without making any graphs etc, how can someone figure out the values of a, b and c?

 

[HallsofIvy]

If x= 4, and y> 0, then y²= 1- 16/25= 9/25 so y= 3/5. If x= -4, then y= 3/5 also, of course, so the two points are (4, 3/25) and (-4, 3/25). In order that the parabola go through both points you must have 16a+ 4b+ c= 3/25 and 16a- 4b+ c= 3/25. Subtracting those two equations immediately gives 8b= 0.

The derivative of y with respect to x is given by (2/24)x+ 2y dy/dx= 0 or dy/dx= (-1/25)(x/y). At (4, 3/25) that is dy/dx= (1/25)(100/3)= 4/3 and at (-4, 3/25) dy/dx= -4/3.

If y= ax²+ bx+ c, then y'= 2ax+ b so you must have 4/3= 2a(-4)+ b and -4/3= -2a+ b. Solve for those for a and b and then use the fact that 16a+ 4b+ c= 3/25 to find c.

 

[roam]

Yes...

For the ellipse: \frac{d}{dx} (\frac{x^2}{25} + y^2) = \frac{2x}{25} + 2y \frac{dy}{dx} = 0

\Rightarrow dy/dx = \frac{-x}{25y}

At x = ±4, \frac{16}{25} + y^2 = 1 so that y = \frac{3}{5}

Then \frac{dy}{dx}\right|_{4, 3/5} = \frac{-4}{15},

\frac{dy}{dx}\right|_{-4, 3/5} = \frac{4}{15}


For the parabola, dy/dx = 2ax + b , so that we want;

slope at (4, 0.6): 8a + b = -4/15

slope at (−4, 0.6): -8a + b = 4/15

and adding these equations gives: a = \frac{-1}{30}

The two curves meet at (4, 0.6), so that y = ax² +c satisfies 3/5 = -16/30 + c

=> c = \frac{34}{30}

The equation of the parabola should be: y = \frac{34 - x^2}{30}

 

[anantchowdhary]

you can also use quadratic equations and form a general equation of a tagent to a standard parabola and also that for an ellipse..Use the condition that the straight line will touch the parabola at one point or ,the point on the parabola which satsfies the line's eqn is one and only one..