Find a Parabola with Vertex (3,1) Focus (4,1)
Find the vertex of 4y^2+4y-16x+13=0
Find the center of an ellipse 9x^2+4y^2-36x-24y-36=0
Find an ellipse with minor axis of 8 and vertices at (-9,3),(7,3)
answers please btw steps arent necessary
1. V(3 , 1), F(4 , 1).
X = a(Y - k)^2 + h,
VF = 1/4a = 4 - 3 = 1,
1/4a = 1,
4a = 1,
a = 1/4.
Eq: X = 1/4(Y - 1)^2 + 3.
2. 4Y^2 + 4Y - 16X +13 = 0.
-16X + 4Y^2 + 4Y = -13,
-4X + Y^2 + Y = -13/4,
-4X + Y^2 + Y + (1/2)^2 = -13/4 + 1/4.
-4X + ((Y + 1/2)^2 = -12/4 = -3,
-4X = -(Y + 1/2)^2 - 3,
X = 1/4(Y + 1/2)^2 + 3/4.
V(3/4 , -1/2).
To find a parabola with a given vertex and focus, we can use the standard form equation of a parabola: (x-h)^2 = 4p(y-k). In this equation, (h, k) represents the vertex of the parabola, and p represents the distance between the vertex and focus.
In this case, the given vertex is (3,1) and the focus is (4,1). Since the y-coordinate of both the vertex and focus is the same, we know that p = 1.
Using the standard form equation, we have (x-3)^2 = 4(1)(y-1).
Expanding and rearranging the equation, we get x^2 - 6x + 9 = 4y - 4.
Further simplifying, we have x^2 - 6x - 4y + 13 = 0.
Therefore, the equation of the parabola with vertex (3,1) and focus (4,1) is x^2 - 6x - 4y + 13 = 0.
To find the vertex of the equation 4y^2 + 4y - 16x + 13 = 0, we need to rewrite the equation in the standard form of a parabola, (y - k)^2 = 4p(x - h).
First, we'll complete the square for the y terms:
4y^2 + 4y = 16x - 13
4(y^2 + y) = 16x - 13
4(y^2 + y + 1/4) = 16x - 13 + 1
4(y + 1/2)^2 = 16x - 12
Now, we can rewrite in standard form:
(y + 1/2)^2 = 4(x - 3/4)
Comparing this equation with the standard form, we find that the vertex of the parabola is at (h, k) = (3/4, -1/2).
To find the center of an ellipse, we need to rewrite the equation in the standard form of an ellipse, (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) represents the coordinates of the center, and a and b represent the semi-major and semi-minor axes, respectively.
Given the equation 9x^2 + 4y^2 - 36x - 24y - 36 = 0, we need to complete the squares for both the x and y terms.
9(x^2 - 4x) + 4(y^2 - 6y) = 36
9(x^2 - 4x + 4) + 4(y^2 - 6y + 9) = 36 + 9(4) + 4(9)
9(x - 2)^2 + 4(y - 3)^2 = 99
Now, we can rewrite in standard form:
(x - 2)^2/11 + (y - 3)^2/24 = 1
Comparing this equation with the standard form, we find that the center of the ellipse is at (h, k) = (2, 3).
To find an ellipse with a minor axis of 8 and vertices at (-9,3),(7,3), we can use the standard form equation of an ellipse: (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) represents the coordinates of the center, and a and b represent the semi-major and semi-minor axes, respectively.
Given the vertices (-9,3) and (7,3), we can find the length of the minor axis, which is 2b. In this case, 2b = 8, so b = 4.
Since the ellipse is horizontally aligned, we have the center (h, k) at the midpoint of the vertices, which is ((-9+7)/2,3) = (-1,3).
Now, we have the center of the ellipse at (-1,3) and the semi-minor axis b = 4.
To find the value of a, we can use the distance formula between the center and one of the vertices: sqrt((x2 - x1)^2 + (y2 - y1)^2).
Using (-1,3) as the center and either of the vertices, we have:
a = sqrt((-1 - (-9))^2 + (3 - 3)^2) = sqrt(64 + 0) = 8.
Now, we can write the equation in standard form:
(x + 1)^2/64 + (y - 3)^2/16 = 1.
Therefore, an ellipse with a minor axis of 8 and vertices at (-9,3) and (7,3) has an equation of (x + 1)^2/64 + (y - 3)^2/16 = 1.