Question

1. (0 points) find the center, vertices, and foci of the ellipse with equation:

\\x^2 + 16y^2 + 6x + 64y + 57 = 0\\

and graph the ellipse.

center:
vertices:
foci:

Explanation:

Step 1: Group the terms

Group the \(x\) and \(y\) terms together.

$$(x^2 + 6x) + (16y^2 + 64y) = -57$$

Step 2: Factor out coefficients

Factor out the coefficient of \(y^2\).

$$(x^2 + 6x) + 16(y^2 + 4y) = -57$$

Step 3: Complete the square

Add necessary constants to complete the squares.

$$(x^2 + 6x + 9) + 16(y^2 + 4y + 4) = -57 + 9 + 64$$

Step 4: Simplify the equation

Write in perfect square form and simplify.

$$(x + 3)^2 + 16(y + 2)^2 = 16$$

Step 5: Write in standard form

Divide both sides by 16.

$$\frac{(x + 3)^2}{16} + \frac{(y + 2)^2}{1} = 1$$

Step 6: Identify the center

Find the center \((h, k)\).

$$(h, k) = (-3, -2)$$

Step 7: Find the vertices

Use \(a = 4\) along the horizontal axis.

$$(-3 \pm 4, -2) \Rightarrow (-7, -2) \text{ and } (1, -2)$$

Step 8: Find the foci

Calculate \(c = \sqrt{a^2 - b^2}\) and locate foci.

$$c = \sqrt{16 - 1} = \sqrt{15} \Rightarrow (-3 \pm \sqrt{15}, -2)$$

Answer:

Center: \((-3, -2)\)
Vertices: \((-7, -2)\) and \((1, -2)\)
Foci: \((-3 - \sqrt{15}, -2)\) and \((-3 + \sqrt{15}, -2)\)