1.) Write the standard form equation for the following Ellipse. Use the information given below.
Foci: (0,0) and (4,0)
2.) Write the standard form equation for the following Hyperbola. Use the information below.
Vertices (-2,1) and (2,1)
Foci: (-3,1) and (3,1)
Not enough information for 1.
all we know is that the centre is (2,0) and c = 2
(x-2)^2 /a^2 + y^2 / b^2 = 1
where a^2 = b^2 + 4
2. centre is (0,1) , a = 2 , c = 3
in a hyperbola: a^2 + b^2 = c^2
4 + b^2 = 9
b^2 = 5
x^2 /4 - (y-1)^2 /5 = 1
For #1 it says:
Major Axis of length 8
Foci: (0,0) and (4,0)
I forgot to copy the next part. Sorry.
To write the standard form equation for an ellipse or a hyperbola, we need to understand the key elements of these conic sections. Let's start with the definitions:
1) Ellipse: An ellipse is a curve that forms a closed loop, like a flattened circle. It is defined by two foci (plural for "focus") and the sum of the distances from any point on the ellipse to the two foci remains constant.
2) Hyperbola: A hyperbola consists of two separate curves that look like mirror images of each other. It is defined by two foci, and the difference of the distances from any point on the hyperbola to the two foci is constant.
Now, let's work on each problem step by step:
1) Equation for the Ellipse with Foci (0,0) and (4,0):
The general standard form equation for an ellipse centered at the origin is:
(x^2 / a^2) + (y^2 / b^2) = 1
where "a" represents the length of the semi-major axis, and "b" represents the length of the semi-minor axis.
In this case, the foci are (0,0) and (4,0).
To find the values of "a" and "b," we use the relationship between the foci and the semi-major axis (2a). In this ellipse, the distance between the foci is 4 units, so the length of the semi-major axis is 4 / 2 = 2 units.
The semi-minor axis "b" can be found using the relationship:
c^2 = a^2 - b^2
where "c" represents the distance between the center and each focus. In this case, c = 2 units (half the distance between the foci), and a = 2 units. By rearranging the formula, we can solve for "b":
c^2 = a^2 - b^2
2^2 = 2^2 - b^2
4 = 4 - b^2
b^2 = 0
b = 0
Since b = 0, the ellipse degenerates into a line (y = 0), which means it is a special case where the semi-minor axis collapses to zero. Thus, the equation for this ellipse becomes:
(x^2 / 2^2) + (y^2 / 0^2) = 1
(x^2 / 4) + (y^2 / 0) = 1
Simplifying further:
x^2 / 4 = 1
x^2 = 4
x = ±2
Therefore, the equation for this ellipse is x = ±2.
2) Equation for the Hyperbola with Vertices (-2,1) and (2,1) and Foci (-3,1) and (3,1):
The general standard form equation for a hyperbola centered at the origin is:
(x^2 / a^2) - (y^2 / b^2) = 1
where "a" represents the distance from the center to each vertex, and "b" represents half the distance between the two branches of the hyperbola.
In this case, the vertices are (-2,1) and (2,1). The distance from the center to each vertex (a) is 2 units.
The distance between the foci, c, can be found using the relationship:
c^2 = a^2 + b^2
In this hyperbola, the foci are (-3,1) and (3,1), and a = 2. By substituting these values into the equation, we can solve for "b":
c^2 = a^2 + b^2
(6^2) = (2^2) + b^2
36 = 4 + b^2
b^2 = 36 - 4
b^2 = 32
b = √32
The equation for this hyperbola becomes:
(x^2 / 2^2) - (y^2 / (√32)^2) = 1
(x^2 / 4) - (y^2 / 32) = 1
Simplifying further:
32x^2 - 4y^2 = 128
8x^2 - y^2 = 32
Therefore, the equation for this hyperbola is 8x^2 - y^2 = 32.