A variable tangent to the ellipse (x/a)^2 +(y/b)^2 meets the parabola y^2=4ax at L and M. Find the locus of the midpoint of LM.

Question

Reiny · Answer

Wow, as Steve pointed out to you yesterday for this same question, this is some serious and messy algebra.

I attempted it with what looked like straightforward actual equations, and it got bad very quickly.

With Wolfram helping me to check for correct in-between steps and avoid same heavy arithmetic ...

Let the ellipse be: x^2/25 + y^2/16 = 1, and the parabola : y^2 = 8x

x^2/25 + y^2/16 = 1
16x^2 + 25y^2 = 400
dy/dx = -16x/25y

I picked y = 3 on the ellipse
then 16x^2 + 225 = 400
x^2 = 175/16
x = ± 5√7/4

I picked the point (5√7/4 , 3)
slope at that point = -16(5√7/4) / 3 = -4√7/15

so the equation of the tangent is
y - 3 = (-4√7/15)(x - 5√7/4)
15y - 45 = -4√7(x - 5√7/4)
15y - 45 = -4√7 x + 35
y = (80 - 4√7 x)/15 **

at this point I used Wolfram to graph the ellipse, the parabola and my calculated tangent.
http://www.wolframalpha.com/input/?i=plot+x%5E2%2F25+%2B+y%5E2%2F16%3D1,y-3+%3D+(-4%E2%88%9A7%2F15)(x-5%E2%88%9A7%2F4),y%5E2%3D8x

Notice the tangent touches the ellipse and crosses the parabola at two points.
So now to find those two points, sub ** into my parabola

[ (80 - 4√7 x)/15 ]^2 = 8x
(80 - 4√7 x)^2 = 1800x
6400 - 640√7 x + 112x^2 = 1800x

112x^2 - x(640√7 + 1800)+ 6400 = 0
didn't feel like doing this, so again used Wolfram
http://www.wolframalpha.com/input/?i=solve+112x%5E2+-+x(640%E2%88%9A7+%2B+1800)+%2B+6400+%3D+0
clicking on 'approximate form'
to get x = 1.9546 and x = 29.235

once again using Wolfram to get
http://www.wolframalpha.com/input/?i=solve+y+%3D+(+(-4%E2%88%9A7)+x+%2B+80)%2F15+for+x+%3D+1.9546
and
http://www.wolframalpha.com/input/?i=solve+y+%3D+(+(-4%E2%88%9A7)+x+%2B+80)%2F15+for+x+%3D+29.235

for the intersection points of
(1.9546,3.9543) and (29.235, -15.2929)

All that remains is to find the trivial calculation of the midpoint for these intersection points.

notice that my Wolfram graph of all three relations verifies this.

So.... if you have the energy and patience to repeat these steps with the general equations, go ahead and have fun. This particular case was enough for me.