1. Identify the conic section represented by: 9y^2+4x^2 - 108y+24x= -144?

Is it ellipse.

2. Find the coordinates of the vertex and the equation of the axis of symmetry for the parabola represented by: x^2 + 4x - 6y + 10 = 0.

vertex: (-2 , 1)
axis of symmetry: x = -2

This is really old

1.

9y^2+4x^2 - 108y+24x= -144
9(y-6)²-9*6² + 4(x+3)²-4*3² = -144
9(y-6)² + 4(x+3)²= -144 + 324 + 36
9(y-6)² + 4(x+3)²= (6√6)²
((y-6)/(2√6))² + ((x+3)/(3√6))²= 1
Does that ring a bell?

2.
This is done by completing the square:
x^2 + 4x - 6y + 10 = 0.
y=(x^2 + 4x + 10)/6
=(1/6)(x+2)²+1
=a(x-h)²+k (a=1/6, h=-2, k=1)
(h,k) is the vertex.
The equation of the axis of symmetry is
x=h

identify the conic

9 y^2+4y^2-108y+24x=-144, determine the center

i think it is a hyperbola

nvm that last post, i thought this was hw help

1. To identify the conic section represented by the equation 9y^2 + 4x^2 - 108y + 24x = -144, we need to rearrange the equation into a standard form.

First, let's group the terms with y and x separately:

9y^2 - 108y + 4x^2 + 24x = -144

Next, let's complete the square for the y-terms and x-terms separately.

For the y-terms:

9(y^2 - 12y) = -4x^2 - 24x - 144

We can complete the square by adding (half of the coefficient of y)^2 to both sides:

9(y^2 - 12y + 36) = -4x^2 - 24x - 144 + 324

Simplifying further:

9(y - 6)^2 = -4x^2 - 24x + 180

Let's do the same for the x-terms:

4(x^2 + 6x) = -9(y - 6)^2 + 180

By completing the square:

4(x^2 + 6x + 9) = -9(y - 6)^2 + 180 + 36

Simplifying:

4(x + 3)^2 = -9(y - 6)^2 + 216

Now we have the equation in standard form, which is:

4(x + 3)^2 / 216 = -9(y - 6)^2 / 216 + 1

Comparing this equation to the standard form equation for conic sections, we can see that it represents an ellipse. The coefficients of x^2 and y^2 have the same sign, and the denominator is greater than the numerator.

Therefore, the conic section represented by the given equation is an ellipse.

2. To find the coordinates of the vertex and the equation of the axis of symmetry for the parabola represented by x^2 + 4x - 6y + 10 = 0, we can start by rearranging the equation into the standard form.

x^2 + 4x = 6y - 10

Let's complete the square for the x-terms:

(x^2 + 4x + 4) = 6y - 10 + 4

(x + 2)^2 = 6y - 6

Now, the equation is in the form (x - h)^2 = 4p(y - k), where (h, k) represents the vertex of the parabola.

Comparing the equation to the standard form, we can identify that the vertex, (h, k), is (-2, -1/6). Therefore, the coordinates of the vertex are (-2, -1/6).

The axis of symmetry for a parabola is a vertical line that passes through its vertex. In this case, the axis of symmetry is the line x = -2.

Therefore, the coordinates of the vertex are (-2, -1/6), and the equation of the axis of symmetry is x = -2.