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.
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.
To find the locus of the midpoint of LM, let's start by understanding the given problem:
We have an ellipse with the equation (x/a)^2 + (y/b)^2 = 1, where a and b are constants.
We also have a parabola with the equation y^2 = 4ax, where a is a constant.
We are given that there is a variable tangent to the ellipse that intersects the parabola at points L and M.
Let's proceed step by step to find the locus of the midpoint of LM:
Step 1: Find the equation of the tangent to the ellipse
To find the equation of the tangent, we need to differentiate the equation of the ellipse implicitly with respect to x:
2(x/a) + 2(y/b) * (dy/dx) = 0
Simplifying, we can express dy/dx as a function of x and y:
(dy/dx) = - (b/a) * (x/y)
Step 2: Find the coordinates of the points L and M on the parabola
Since the equation of the parabola is given as y^2 = 4ax, we can substitute the value of y from the equation of the tangent into the equation of the parabola:
(y/b) ^ 2 = 4a * (x/a)
Simplifying, we get:
y^2 / b^2 = 4a * (x/a)
y^2 = 4ab^2 * x
Step 3: Find the coordinates of the midpoint of LM
Let's call the coordinates of L as (x1, y1) and the coordinates of M as (x2, y2).
The midpoint of LM can be found by taking the average of the x-coordinates and the y-coordinates:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Step 4: Find the locus of the midpoint
Now, substituting the coordinates of L and M into the equation for the midpoint, we get:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
= ((x1 + x2)/2, (- (b/a) * (x1/y1) + (- (b/a) * (x2/y2))/2)
Simplifying further:
Midpoint = ((x1 + x2)/2, (-b/a) * ((x1/y1) + (x2/y2))/2)
= ((x1 + x2)/2, (-b/a) * (x1 * y2 + x2 * y1)/(2 * y1 * y2))
The locus of the midpoint is the set of points that satisfy this equation.
This is the final answer - the locus of the midpoint of LM in terms of the variables x1, x2, y1, and y2.