Find the vertices and foci of this ellipse 4x^2+16y^2+64x+64y+256=0
4x^2+16y^2+64x+64y+256=0
4x^2+64x+256 + 16y^2+64y+64 = -256
4(x+8)^2 + 16(y+2)^2 = -256 + 256 + 64
(x+8)^2 + 4(y+2)^2 = 16
(x+8)^2/16 + (y+2)^2/4 = 1
h=-8 k=-2
a=4 b=2 c=√12
center: (-8,-2)
vertices: (-12,-2) (-4,-2)
foci: (-8-√12,-2) (-8+√12,-2)
To find the vertices and foci of the given ellipse, we need to rewrite the equation in the standard form:
( x - h )^2 / a^2 + ( y - k )^2 / b^2 = 1
where (h, k) represents the center of the ellipse, and 'a' and 'b' represent the length of the major and minor axes, respectively.
Let's start by completing the square for both the x and y terms in the given equation.
4x^2 + 16y^2 + 64x + 64y + 256 = 0
Dividing the entire equation by 4 to make the coefficient of x^2 equal to 1:
x^2 + 4y^2 + 16x + 16y + 64 = 0
Now, we can complete the square for the x terms:
x^2 + 16x + 64 + 4y^2 + 16y = 0
(x + 8)^2 + 4y^2 + 16y = 0
To complete the square for the y terms, we need to factor out a 4 from the y^2 and y terms:
(x + 8)^2 + 4(y^2 + 4y) = 0
To form a perfect square for y, we add (4/2)^2 = 4 to both sides of the equation:
(x + 8)^2 + 4(y^2 + 4y + 4) = 0 + 4
(x + 8)^2 + 4(y + 2)^2 = 4
Dividing through by 4:
[(x + 8)^2] / 4 + [(y + 2)^2] / 1 = 1
Now we have the equation in standard form. Comparing this with the standard equation, we can see that:
h = -8
k = -2
a^2 = 4 (a = 2)
b^2 = 1 (b = 1)
Thus, the center of the ellipse is (-8, -2), and the major axis has a length of 2, while the minor axis has a length of 1.
To find the vertices, we need to identify the points where the major axis intersects the ellipse. In this case, the major axis is horizontal due to the coefficient of (x - h)^2 being larger than the coefficient of (y - k)^2.
The vertices are located at (h ± a, k), which in this case is (-8 ± 2, -2). Therefore, the vertices are (-10, -2) and (-6, -2).
To find the foci of the ellipse, we need to use the formula:
c = √(a^2 - b^2)
In this case, c = √(4 - 1) = √3 ≈ 1.732.
The foci are located at (h ± c, k), so the foci are approximately (-8 - 1.732, -2) and (-8 + 1.732, -2). Hence, the foci are approximately (-9.732, -2) and (-6.268, -2).
In summary:
Center: (-8, -2)
Vertices: (-10, -2) and (-6, -2)
Foci: (-9.732, -2) and (-6.268, -2)