Final answer:
The vertices of the ellipse are (±2√3, 0) and the foci are (0, ±3).
Step-by-step explanation:
The given equation is a equation of an ellipse in standard form:
4x² - 16y² − 32y = 48
To find the vertices and foci of the ellipse, we need to rewrite the equation in a standard form:
4(x - h)² + 16(y - k)² = 48
Dividing both sides of the equation by 48 gives:
(x - h)²/12 + (y - k)²/3 = 1
Comparing the equation with the standard form of an ellipse, we can see that the center of the ellipse is at the point (h, k).
Now, let's find the values of h and k. To do that, we need to complete the square for both x and y terms.
Completing the square for x terms:
(x - h)²/12 = 1
(x - h)² = 12
Hence, h = 0. The x-coordinate of the center is 0.
Completing the square for y terms:
(y - k)²/3 = 1
(y - k)² = 3
Hence, k = 0. The y-coordinate of the center is 0.
Therefore, the center of the ellipse is at the point (0, 0).
The distance from the center to the vertices is equal to the value of 'a'. From the equation, we can see that a = √12 = 2√3.
Therefore, the vertices of the ellipse are (±2√3, 0).
The distance from the center to the foci is equal to the value 'c'. From the equation, we can see that c = √(12-3) = √9 = 3.
Therefore, the foci of the ellipse are (0, ±3).