Find the equation of an ellipse that have eccentricity 2/5 and foci (-1, 3) and (3, 3) .
distance from (-1,3) to (3,3) is 4, so 2c = 4
c = 2
for your ellipse , e = c/a
2/5 = 2/a
a = 5
we also know that a^2 = b^2 + c^2
25 = b^2 + 4
b^2 = √21
the centre is midway between the foci, so the centre is (1,3)
equation:
(x-1)^2 / 25 + (y-3)^2 / 21 = 1
verification:
https://www.wolframalpha.com/input/?i=plot+%28x-1%29%5E2+%2F+25+%2B+%28y-3%29%5E2+%2F+21+%3D+1
notice the a^2 is close to b^2, so it looks almost circular
To find the equation of an ellipse given its eccentricity and foci, we can follow these steps:
Step 1: Find the distance between the foci.
The distance between the foci is given by the formula:
c = √(b^2 - a^2)
where c is the distance between the foci, and a and b are the semi-major and semi-minor axes of the ellipse, respectively.
In this case, the eccentricity (e) is given as 2/5. The eccentricity is related to the lengths of the semi-major and semi-minor axes by the formula:
e = c/a
We can rearrange the formula to solve for c:
c = e * a
Since e = 2/5, we have:
c = (2/5) * a
Step 2: Find the length of the major axis (2a).
The major axis of the ellipse is twice the length of the semi-major axis. We can find it by using the distance formula between the foci (-1, 3) and (3, 3).
Distance between foci = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(3 - (-1))^2 + (3 - 3)^2]
= √[(4)^2 + (0)^2]
= √[16 + 0]
= √16
= 4
We know that c = 4, so:
4 = (2/5) * a
Step 3: Find the value of a.
We can solve the equation 4 = (2/5) * a for a:
4 * 5 = 2a
20 = 2a
a = 10
Step 4: Find the value of b.
Using the equation c = (2/5) * a, we can substitute in the value of a (10) to find b:
c = (2/5) * a
4 = (2/5) * 10
4 = 4
Since the distance between the foci is 4 units and both foci lie on the same horizontal line, this means that the major axis is horizontal. Therefore, a = 10 is the length of the semi-major axis.
Step 5: Write the equation of the ellipse.
Now that we have the values of a and b, we can write the equation of the ellipse in standard form:
For a horizontal ellipse with the major axis on the x-axis:
(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1
The center of the ellipse is the midpoint between the foci, which is (h, k). In this case, the x-coordinate of the center is the average of the x-coordinates of the foci, and the y-coordinate is the same for both foci:
h = (3 - 1) / 2 = 2 / 2 = 1
k = 3
Substituting the values of h, k, a, and b into the equation, we have:
(x - 1)^2 / 10^2 + (y - 3)^2 / 4^2 = 1
Thus, the equation of the ellipse that has an eccentricity of 2/5 and foci (-1, 3) and (3, 3) is:
(x - 1)^2 / 100 + (y - 3)^2 / 16 = 1