what is the vertices of (x+1)^2/4+(y+2)^2/25=1 ?
To determine the vertices of the ellipse given by the equation (x+1)^2/4 + (y+2)^2/25 = 1, you can follow these steps:
Step 1: Identify the center of the ellipse
In the given equation, the center of the ellipse is (-1,-2). This means that the "x" component of the center is -1, and the "y" component is -2.
Step 2: Determine the major and minor axes
The equation of the ellipse can be written as (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 length, and "b" is the semi-minor axis length. Comparing this equation to the given equation, we can identify the values:
- Semi-major axis length (a) = 2 (since a^2 = 4)
- Semi-minor axis length (b) = 5 (since b^2 = 25)
Step 3: Calculate the vertices using the center and axes
For an ellipse with center (h,k), the vertices can be calculated using the following formulas:
Vertex 1: (h-a, k)
Vertex 2: (h+a, k)
Applying these formulas to the given ellipse, we get:
- Vertex 1: (-1-2, -2) = (-3, -2)
- Vertex 2: (-1+2, -2) = (1, -2)
Therefore, the vertices of the ellipse are (-3, -2) and (1, -2).