9x^2+16y^2+64y+28=0

I am supposed to convert this to standard form, but my answer doesn't make sense. One part was 9x^2/92...does that sound right?

16 y^2 + 64 y = - 9 x^2-28

y^2 + 4 y = (- 9 x^2-28 )/16
y^2 + 4 y + 4 = (- 9 x^2-28 )/16 +64/16
(y+2)^2 = (-9 x^2 + 36 )/16
(y+2)^2 + (9/16) x^2 = 36/16
16/36 (y+2)^2 + 9/36 x^2 = 1

x^2/2^2 + (y+2)^2/(3/2)^2 = 1
ellipse with center at (0,-2)
x half axis length 2
y half axis length = 1.5

Shouldn't it equal zero anyway because what does Y equal?

To convert the given equation to standard form, we need to rearrange the terms to isolate the variables on one side and have a constant on the other side. The standard form of an equation of a conic section is generally expressed as:

Ax^2 + By^2 + Cx + Dy + E = 0

In your case, the equation is 9x^2 + 16y^2 + 64y + 28 = 0.

To begin, let's group the x-terms and the y-terms together:

9x^2 + (16y^2 + 64y) + 28 = 0

Next, let's complete the square for the y-terms. To do this, we need to take half of the coefficient of y and square it. Here, we have a coefficient of 64 for y, so:

(16(y^2 + 4y)) = (16(y^2 + 4y + 4)) - 16(4)

Expanding and simplifying this, we get:

16(y^2 + 4y + 4 - 4) = 16(y^2 + 4y) = 16(y + 2)^2 - 16(4)

Now, we can substitute this back into the original equation:

9x^2 + 16(y + 2)^2 - 16(4) + 28 = 0

Expanding and simplifying further, we obtain:

9x^2 + 16(y + 2)^2 - 64 + 28 = 0

9x^2 + 16(y + 2)^2 - 36 = 0

Finally, we can divide the entire equation by the constant term on the right side to simplify it further:

9x^2/36 + (y + 2)^2/9 - 1 = 0

In standard form, the equation is:

x^2/4 + (y + 2)^2/9 - 1/4 = 0

So, it seems there was an error in your answer. The correct expression for 9x^2 in standard form after simplification is x^2/4.