To find the standard form equation of the ellipse and locate the foci, let's analyze the information given.
Step 1: Identify the Components of the Ellipse
The vertical ellipse has:
- Center at \((-2, -3)\)
- The points it passes through include:
- \((2, -3)\) (horizontal radius)
- \((-2, -9)\) (vertical radius)
- \((-6, -3)\) (horizontal radius)
- \((-2, 3)\) (vertical radius)
Step 2: Determine the Radii
Horizontal Radius (a):
- The distance from the center to the point \((2, -3)\) (which is horizontally aligned with the center) is: \[ |2 - (-2)| = 4 \quad \text{(right side)} \] Thus, \(a = 4\).
Vertical Radius (b):
- The distance from the center to the point \((-2, -9)\) (which is vertically aligned with the center) is: \[ |-9 - (-3)| = 6 \quad \text{(downward direction)} \] Thus, \(b = 6\).
Step 3: Write the Standard Form Equation of the Ellipse
Since it is a vertical ellipse, the standard form is: \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \] where:
- \((h, k)\) is the center
- \(a\) and \(b\) are the semi-major and semi-minor axes respectively.
Substituting \(h = -2\), \(k = -3\), \(a = 4\), and \(b = 6\): \[ \frac{(x + 2)^2}{4^2} + \frac{(y + 3)^2}{6^2} = 1 \] \[ \frac{(x + 2)^2}{16} + \frac{(y + 3)^2}{36} = 1 \]
Step 4: Finding the Foci
The distance \(c\) from the center to each focus can be found using the relationship: \[ c = \sqrt{b^2 - a^2} \] Calculating \(c\): \[ c = \sqrt{6^2 - 4^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \]
Step 5: Locate the Foci
The foci for a vertical ellipse are located at: \[ (h, k \pm c) = (-2, -3 \pm 2\sqrt{5}) \] Thus, the coordinates of the foci are: \[ (-2, -3 + 2\sqrt{5}) \quad \text{and} \quad (-2, -3 - 2\sqrt{5}) \]
Conclusion
The standard form equation of the ellipse is: \[ \frac{(x + 2)^2}{16} + \frac{(y + 3)^2}{36} = 1 \] And the foci are: \[ (-2, -3 + 2\sqrt{5}) \quad \text{and} \quad (-2, -3 - 2\sqrt{5}) \]
Answer Choices
- The correct response is: \[ \frac{(x + 2)^2}{16} + \frac{(y + 3)^2}{36} = 1, \quad (-2, -3 \pm 2\sqrt{5}) \]