Questions: Complete the square to determine whether the graph of the equation is an ellipse, a parabola, a hyperbola, a degenerate conic, or has no solution. 3x^2+4y^2-18x-8y+31=0 ellipse parabola hyperbola degenerate conic no solution If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the separated list of equations. If an answer does not exist, enter DNE.) center (x, y)=( ) focus (x, y)=( DNE focus (x, y)=( smaller x-value or only focus) vertex (x, y)=( ) (larger x-value) vertex (x, y)=( ) (larger x-value)

Solution Steps

To determine the type of conic section represented by the given equation, we need to complete the square for both the $x$ and $y$ terms. This will help us rewrite the equation in a standard form that can be easily identified as an ellipse, parabola, hyperbola, or degenerate conic. Once the equation is in standard form, we can identify the conic section and, if applicable, find its properties such as center, foci, vertices, and axes lengths.

The given equation is:

$$3x^2 + 4y^2 - 18x - 8y + 31 = 0$$

First, we rearrange the terms to group the $x$ and $y$ terms:

$$3x^2 - 18x + 4y^2 - 8y = -31$$

For the $x$ terms, factor out the coefficient of $x^2$:

$$3(x^2 - 6x) + 4y^2 - 8y = -31$$

Complete the square inside the parentheses:

1. Take half of the coefficient of $x$, which is $-6$, giving $-3$.
2. Square it to get $9$.
3. Add and subtract $9$ inside the parentheses:

$$3(x^2 - 6x + 9 - 9) + 4y^2 - 8y = -31$$

This simplifies to:

$$3((x-3)^2 - 9) + 4y^2 - 8y = -31$$

Distribute the $3$:

$$3(x-3)^2 - 27 + 4y^2 - 8y = -31$$

For the $y$ terms, factor out the coefficient of $y^2$:

$$3(x-3)^2 + 4(y^2 - 2y) = -31 + 27$$

Simplify the right side:

$$3(x-3)^2 + 4(y^2 - 2y) = -4$$

Complete the square inside the parentheses for $y$:

1. Take half of the coefficient of $y$, which is $-2$, giving $-1$.
2. Square it to get $1$.
3. Add and subtract $1$ inside the parentheses:

$$3(x-3)^2 + 4((y-1)^2 - 1) = -4$$

This simplifies to:

$$3(x-3)^2 + 4(y-1)^2 - 4 = -4$$

Distribute the $4$:

$$3(x-3)^2 + 4(y-1)^2 - 4 = -4$$

Add $4$ to both sides:

$$3(x-3)^2 + 4(y-1)^2 = 0$$

The equation $3(x-3)^2 + 4(y-1)^2 = 0$ suggests that both squared terms must be zero for the equation to hold, which implies:

$$(x-3)^2 = 0 \quad \text{and} \quad (y-1)^2 = 0$$

Thus, $x = 3$ and $y = 1$. This is a single point, indicating a degenerate conic.

Final Answer