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).
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.
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
b^2+c^2 = a^2
A) To find the equation of an ellipse, we use the standard form of the equation:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
where (h,k) is the center of the ellipse, 'a' is the semi-major axis, and 'b' is the semi-minor axis.
In this case, the center of the ellipse is given as c(-2,3). The vertex is given as v(-2,10), which means the distance between the center and vertex is the value of 'a'. We also have one focus at f(-2,7), which means the distance between the center and focus is the value 'c'.
Using the formula c^2 = a^2 - b^2, we can determine the value of 'b'. Since the focus is above the center, 'c' is the difference in y-coordinates between them, which gives c = 7 - 3 = 4.
Now, we have c = 4, a = 10 - 3 = 7. Using the value of 'c', we can determine the value of 'b' as follows:
c^2 = a^2 - b^2
4^2 = 7^2 - b^2
16 = 49 - b^2
b^2 = 49 - 16
b = √33
Finally, we can plug the values into the equation:
(x + 2)^2/49 + (y - 3)^2/33 = 1
Therefore, the equation of the ellipse is (x + 2)^2/49 + (y - 3)^2/33 = 1.
B) In this case, we have the center at c(-1,2), focus at F(-1,5), and a point on the ellipse at P(5,6). We need to determine the values of 'a' and 'b' to write the equation of the ellipse.
The distance between the center and the focus is 'c', while the distance between the center and the given point on the ellipse is 'r'. The equation for the distance between two points is:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
First, let's find the value of 'c':
c = √[(5 - (-1))^2 + (6 - 2)^2]
c = √[(6)^2 + (4)^2]
c = √[36 + 16]
c = √[52]
c = 2√13
Now, let's find the value of 'r':
r = √[(5 - (-1))^2 + (6 - 2)^2]
r = √[(6)^2 + (4)^2]
r = √[36 + 16]
r = √[52]
r = 2√13
Since 'a' is the distance between the center and the point on the ellipse, we have:
a = r = 2√13
Next, we can use the formula c^2 = a^2 - b^2 to determine 'b':
c^2 = a^2 - b^2
(2√13)^2 = (2√13)^2 - b^2
52 = 52 - b^2
b^2 = 0
b = 0
Since 'b' is zero, the equation simplifies to:
(x + 1)^2/(2√13)^2 + (y - 2)^2/0^2 = 1
(x + 1)^2/(2√13)^2 = 1
Therefore, the equation of the ellipse is (x + 1)^2/(2√13)^2 = 1.