SOLUTION: The center of an ellipse is on (-2, -1) and one of its vertex is on (3, -1). It the length of each latus rectum is 4, find the equation of the ellipse, its excentricity and the coo

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The center of an ellipse is on (-2, -1) and one of its vertex is on (3, -1). It the length of each latus rectum is 4, find the equation of the ellipse, its excentricity and the coo      Log On


   



Question 1088643: The center of an ellipse is on (-2, -1) and one of its vertex is on (3, -1). It the length of each latus rectum is 4, find the equation of the ellipse, its excentricity and the coordinates of its foci.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1 is an ellipse with center (h,k), horizontal axis length 2a, and vertical axis length 2b,vertices are at ( h± a,k), foci is at ( h± c, k)
Note that the center is the midpoint of the segment joining the foci.

if the center of an ellipse is on (-2, -1), we have h=-2 and k=-1
vertices are at ( -2± a,-1 )=(3, -1)
=>-2%2Ba=3
=>a=5+
foci is at ( h± c, k)= ( -2± c, -1)
so far, your equation is:
%28x%2B2%29%5E2%2Fa%5E2%2B%28y%2B1%29%5E2%2Fb%5E2=1
the length of each latus rectum is: LR=4
LR=2b%5E2%2Fa=>4=2b%5E2%2F5=>20=2b%5E2=>b%5E2=10
c=sqrt%28a%5E2-b%5E2%29=sqrt%2825-10%29=sqrt%2815%29

foci is at:
( -2± sqrt%2815%29, -1)=>the coordinates of its foci
( +1.87, -1) or ( +-5.87, -1)
your equation is:
%28x%2B2%29%5E2%2F%285%29%5E2%2B%28y%2B1%29%5E2%2F10=1
%28x%2B2%29%5E2%2F25%2B%28y%2B1%29%5E2%2F10=1=>the equation of the ellipse
and, its eccentricity: c%2Fa=+sqrt%2815%29%2F50.77