if you have the ellipse (x^2/a^2)+(y^2/b^2)=1 and a>b what would the sum of the distances from a point (x,y) on the ellipse to each of the foci be.
2 answers
going along with the above problem how would you determine two formulae that calculates the distance from a point on the ellipse to each of the foci. please help
by definition , the sum of the distances from a point P(x,y) to each of the focal points is
2a
see http://mathworld.wolfram.com/Ellipse.html
if you want the individual distances, then first find the foci which would be (c,0) and (-c,0) by using
a^2 = b^2 + c^2, for a>b and then using the distance between two points formula
distance = √((x-c)^2 + y^2)) and
distance = √((x+c)^2 + y^2))
e.g.
for x^2/25 + y^2/9 = 1
a=5, b=3, the by 5^2 = 3^2 + c^2
c = +/- 4
so the foci are (4,0) and (-4,0)
a point P could be (2,(√(189)/5))
(I got that by subbing x=2 into my equation and solving for y)
then
distance#1 = √(2-4)^2 + (√(189)/5 - 0)^2) = √(289/25) = 3.4
distance #2 = √(2+4)^2 + (√(189)/5 - 0)^2) = √(1089/25) = 6.6
notice 3.4 + 6.6 = 10 or 2a
2a
see http://mathworld.wolfram.com/Ellipse.html
if you want the individual distances, then first find the foci which would be (c,0) and (-c,0) by using
a^2 = b^2 + c^2, for a>b and then using the distance between two points formula
distance = √((x-c)^2 + y^2)) and
distance = √((x+c)^2 + y^2))
e.g.
for x^2/25 + y^2/9 = 1
a=5, b=3, the by 5^2 = 3^2 + c^2
c = +/- 4
so the foci are (4,0) and (-4,0)
a point P could be (2,(√(189)/5))
(I got that by subbing x=2 into my equation and solving for y)
then
distance#1 = √(2-4)^2 + (√(189)/5 - 0)^2) = √(289/25) = 3.4
distance #2 = √(2+4)^2 + (√(189)/5 - 0)^2) = √(1089/25) = 6.6
notice 3.4 + 6.6 = 10 or 2a