16x^2+16y^2+64x-32y+55=0
Identify the conic and find the center, radius(for circle), and a and b(for ellipse or hyperbola) sketch the graph
first group and partially factor
16(x^2 + 4x + .....) + 16(y^2 - 2y + .....) = -55
complete the squares
16(x^2 + 4x + 4) + 16(y^2 - 2y + 1) = -55 + 64 + 16
16(x+2)^2 + 16(y - 1)^2 = 25
(x + 2)^2 + (y - 1)^2 = 25/16
identify the conic and state its properties
To identify the conic and find the center, radius, and the values of a and b (if applicable), we need to manipulate the given equation into a standard form. Let's start by completing the square for both the x and y terms.
16x^2 + 16y^2 + 64x - 32y + 55 = 0
Rearranging the equation:
16x^2 + 64x + 16y^2 - 32y + 55 = 0
Now, let's complete the squares by adding and subtracting appropriate constants:
16(x^2 + 4x) + 16(y^2 - 2y) + 55 = 0
Completing the square for the x terms:
16(x^2 + 4x + 4 - 4) + 16(y^2 - 2y) + 55 = 0
16((x + 2)^2 - 4) + 16(y^2 - 2y) + 55 = 0
16(x + 2)^2 - 64 + 16(y^2 - 2y) + 55 = 0
16(x + 2)^2 + 16(y^2 - 2y) - 9 = 0
Completing the square for the y terms:
16(x + 2)^2 + 16(y^2 - 2y + 1 - 1) - 9 = 0
16(x + 2)^2 + 16((y - 1)^2 - 1) - 9 = 0
16(x + 2)^2 + 16(y - 1)^2 - 16 - 9 = 0
16(x + 2)^2 + 16(y - 1)^2 - 25 = 0
Now that we have obtained the equation in standard form, we can identify the conic.
The equation is that of an ellipse since both the x^2 and y^2 terms have the same coefficient with the same sign.
The center of the ellipse is given by (-h, -k), where (h, k) is the coordinates of the center. From the equation, we can see that the center is (-2, 1).
The general form of the equation for an ellipse is:
[(x - h)^2 / a^2] + [(y - k)^2 / b^2] = 1
Comparing this to our equation, we can see that a^2 is equal to 1/16 and b^2 is equal to 1/16.
Therefore, the radius of the ellipse is the square root of 1/16, which is 1/4.
Now, let's sketch the graph. The center is at (-2, 1), and the radius is 1/4.