Find the center, vertices, and foci of the ellipse with equation 3x^2 + 8y^2 = 24.
first place in standard form:
x^2/8 + y^2/3 = 1
Since a^2 = 8 is greater than b^2 = 3, the major axis is horizontal. So, we have
a^2 = 8
b^2 = 3
c^2 = a^2-b^2 = 5
Now just read off the info you want:
center: (0,0)
foci: (±c,0)
vertices: (±a,0)
To find the center, vertices, and foci of an ellipse, we can start by rearranging the equation of the ellipse into standard form:
Divide both sides of the equation by 24:
(3x^2)/24 + (8y^2)/24 = 1
Simplify:
x^2/8 + y^2/3 = 1
From the standard form, we can determine the important values needed to find the center, vertices, and foci of the ellipse. The standard form of an ellipse is:
(x - h)^2/a^2 + (y - k)^2/b^2 = 1
The center of the ellipse (h, k) is (0, 0) because there are no terms with x or y inside parentheses.
Now, let's find the values of a and b. The value of a represents the distance between the center and the vertex along the x-axis, while the value of b represents the distance between the center and the vertex along the y-axis.
From the equation, we can see that a^2 = 8 and b^2 = 3. Therefore, a = √8 = 2√2 and b = √3.
So, the vertices of the ellipse are located at (±2√2, 0).
To find the foci of the ellipse, we can use the formula c^2 = a^2 - b^2, where c represents the distance from the center to the foci.
Plugging in the values, we have c^2 = (2√2)^2 - √3^2
c^2 = 8 - 3 = 5
c = √5
Therefore, the foci of the ellipse are located at (±√5, 0).
In summary:
- Center: (0, 0)
- Vertices: (±2√2, 0)
- Foci: (±√5, 0)