Show that for any point P = (c, d) on the ellipse given by x^2 + 4y^2 = 1...

Question

and the sum of the distances 9 from these points to the two foci is 6, namely (F1)P + (F2)P = 6, where F1 and F2 are the two foci of this ellipse.

Answer 1

x^2 + 4y^2 = 1 is an ellipse with major (longer) axis on the x axis

x^2/1^2 + y^2/(1/2)^2 = 1

where a =1 and b =1/2, the two semi-axes of the ellipse

c^2 = a^2 - b^2 = 1 -1/-4= 3/4

c = + or -sqr3/2 = the x coordinates of the 2 foci (-sqr3/2, 0), (sqr3/2, 0)

distance from (x,y) on the ellipse to each focus, added together always = 2, not 6

foci are (sqr3/2, 0), (-sqr3/2, 0), both on the x axis

try the point (c,d)=(x,y) = (1,0)

it's 1+sqr3/2 from one focus and 1-sqr3/2 from the second focus. Add them to get 2, Not 6

try the largest point on the ellipse, (0, 1/2) top of the vertical axis

it's d^2 = (sqr3/2)^2 + (1/2)^2 = 3/4 +1/4 =1

for distance to one focus

multiply by two to get sum of distances to both foci = 2(1) = 2

there must be a misprint or mis-copy of the problem details somewhere