Graph the equation. Identify the vertices, co-vertices, and foci of the ellipse.

64x^2+16y^2=1024

This is what i have so far
vertices (0,8)(0,-8) foci(0,4√3),(0,-4√3)

stop homework dumping

Sorry, that wasn't one of the choices!!!! Hopefully someone who knows their algebra can help me answer it correctly :-)

To graph the equation 64x^2 + 16y^2 = 1024, we need to rewrite it in the standard form for an ellipse. The standard form equation for an ellipse is:

(x-h)^2/a^2 + (y-k)^2/b^2 = 1

Where (h,k) represents the center of the ellipse, a represents the distance from the center to the vertices, and b represents the distance from the center to the co-vertices.

Given equation: 64x^2 + 16y^2 = 1024

First, we divide the entire equation by 1024 to simplify it:

(64x^2)/1024 + (16y^2)/1024 = 1024/1024

Simplify further:

x^2/16 + y^2/64 = 1

Now we can see that the equation is in the standard form of an ellipse. The vertex is located at the center of the ellipse, so (h,k) = (0,0), meaning the center of the ellipse is at the origin (0,0).

The distance from the center to the vertices is represented by 'a', which is equal to the square root of the denominator of x^2 in the standard form equation. In this case, a = √16 = 4.

Therefore, the vertices will be at (0,±4).

The distance from the center to the co-vertices is represented by 'b', which is equal to the square root of the denominator of y^2 in the standard form equation. In this case, b = √64 = 8.

Therefore, the co-vertices will be at (±8,0).

To find the foci, we need to find the value of 'c'. In the standard form equation, c is equal to the distance from the center to the foci and is calculated using the formula c = √(a^2 - b^2).

In this case, a^2 = 4^2 = 16 and b^2 = 8^2 = 64. Therefore, c = √(16 - 64) = √(-48).

Since c is imaginary, it means that the ellipse does not have any foci.

So, to summarize:

Vertices: (0, 4) and (0, -4)
Co-vertices: (8, 0) and (-8, 0)
No foci for this ellipse.