y=2x+c is a chord of the ellipse x^2+4y^2=16. as c varies find the locus of the midpoint of the chord
ellipse: x^2 + 4y^2 = 16
chord: y = 2x + c
Points of intersection
Substituting for y = 2x + c into the eqn for the ellipse
x^2 + 4(2x + c)^2 = 16
x^2 + 16x^2 + 16cx + 4c^2 = 16
17x^2 + 16cx + 4c^2 – 16 = 0
Solving for x, using the quadratic formula,
x = {-16c ± √(256c^2 – 4.17.(4c^2 – 16)) } / (34)
x = {-16c ± √(256c^2 – 272c^2 + 1088)) } / (34)
x = {-16c ± √(1088 – 16c^2)) } / (34)
x = {-16c ± 4√(68 – c^2)) } / (34)
Let x = a ± b, where a = -16c/34, and b = 4√(68 – c^2)/34
Then, x1 = a – b, x2 = a + b
And, y1 = 2(a – b) + c, y2 = 2(a + b) + c
The midpoint, P is P(r, s) = ((x1+x2)/2, (y1+y2)/2)
(r, s) = (a, 2a + c)
Substituting for a = -8c/17,
P(r, s) = (-8c/17, -16c/17 + c)
P(r, s) = (-8c/17, c/17), giving
r = -8c/17, s = c/17
Eliminating c from the two eqns,
r = -8(17s)/17
r = -8s
Since r is the x-coordinate, and s is the y-coordinate, then we can write the locus as,
x = -8y, or
Locus is: y = -x/8