Question

Write the equation of ellipse that satisfy the following condition A)centre at c(-2,3),one focus is f(-2,7) and one vertex is v(-2,10). B) centr at c(-1,2) focus at F(-1,5) and passes through point P(5,6).

Answer 1

A)
centre at (-2,3)
--> (x+2)^2 /a^2 + (y-3)^2 / b^2 = 1
focus at (-2,7) implies the ellipse is vertical and
c = 7-3 = 4
vertex is (-2,10) , so a = 10=3 = 7
and a^2 + c^2 = b^2
7^2 + 4^2 = b^2 = 65

(x+2)^2 /49 + (y-3)^2 / 65 = 1


B) centre (-1,2) ---> (x+1)^2/a^2 + (y-2)^2/b^2 = 1
one focus at (-1,5) , so c = 5-2=3
Again, since the ellipse is vertical,
b^2 = a^2 + c^2 = a^2 + 9

equation so far:
(x+1)^2/a^2 + (y-2)^2/(a^2 + 9) = 1
but (5,6) lies on it, so
(5+1)^2/a^2 + (6-2)^2/(a^2 + 9) = 1
36/a^2 + 16/(a^2+9) = 1
times a^2(a^2 + 9)
36(a^2 + 9) + 16a^2 = a^2(a^2 + 9)
36a^2 + 324 + 16a^2 = a^4 + 9a^2
a^4 - 43a^2 - 324 = 0

solve for a^2 using the formula, since it does not factor, rejecting any a^2 = negative
I was expecting a "nicer" solution, so you might want to check my arithmetic.

Answer 2

alternate solution for B:
let's "move" everything to standard position
using (x,y) ---> (x+1, y-2)

centre (-1,2) ---> centre (0,0)
F(-1,5) ----> focus( 0, 3) , so c = 3
P(5,6) -----> P' (6, 4)

b^2 = a^2 + c^2 = a^2 + 9

starting equation:
x^2 /a^2 + y^2/b^2 = 1
x^2/a^2 + y^2/(a^2+9) = 1
but (6,4) lies on it
36/a^2 + 16/(a^2+9) = 1
36a^2 + 324 + 16a^2 = a^4 + 9a^2
a^4 -43a^2 - 324 = 0
same equation as above

btw, the exact values are:
a^2 = (43 + √3145)/2
b^2 = (61 + √3145)/2
I get a^2 = appr 49.54 or a^2 = a negative
then b^2 = appr 58.54

final equation:
(x+1)^2/49.54 + (y-2)^2/58.54 = 1

Answer 3

b^2+c^2 = a^2