1. Find the general equation of a Parabola with Vertex at (2,1) and directrix at x=5.
Since the directrix is a vertical line, the parabola opens horizontally. The focus is located at a distance of a = 3 units to the left of the vertex.
The general equation of a parabola with vertex (h,k), focus (h-a,k), and directrix x = h+a is:
(x - h)^2 = 4a(y - k)
Substituting the given values, we have:
(x - 2)^2 = 12(y - 1)
This is the general form of the parabolic equation. To sketch the graph, we can plot the vertex (2,1) and the point on the directrix that is closest to the vertex, which is (5,1). The focus is located 3 units to the left of the vertex, so it is at (-1,1). We can also plot a few more points on the parabola by substituting values of x and solving for y. The resulting graph should be a horizontal parabola opening to the left.
2. Find the general equation of a Parabola with Focus at (-1, -6) and directrix at y=0.
Since the directrix is a horizontal line, the parabola opens vertically. The vertex is located at the midpoint between the focus and the directrix, which is (−1,−3).
The general equation of a parabola with vertex (h,k), focus (h,k+a), and directrix y = k-a is:
(y - k)^2 = 4a(x - h)
Substituting the given values, we have:
(y + 3)^2 = 24(x + 1)
This is the general form of the parabolic equation. To sketch the graph, we can plot the vertex (−1,−3) and the focus (−1,−6). We can also plot a few more points on the parabola by substituting values of y and solving for x. The resulting graph should be a vertical parabola opening downward.
3. Find the general equation of a Parabola with Vertex at (2,3) and with the end points of Latus Rectum are (6,1) & (-2,1).
The Latus Rectum (LR) is a line segment that passes through the focus and is perpendicular to the axis of symmetry. The length of the LR is equal to 4a, where a is the distance between the vertex and the focus. Since the LR passes through (6,1) and (−2,1), its length is 8 units. Thus, a = 2 units.
The vertex is located at (2,3), which is the midpoint of the LR. Therefore, the focus is located at a distance of 2 units above and below the midpoint, which are (2,5) and (2,1), respectively.
The general equation of a parabola with vertex (h,k), focus (h,k+a), and directrix y = k-a is:
(y - k)^2 = 4a(x - h)
Substituting the given values, we have:
(y - 3)^2 = 16(x - 2)
This is the general form of the parabolic equation. To sketch the graph, we can plot the vertex (2,3) and the focus (2,5) and (2,1). We can also plot the LR by connecting the endpoints (6,1) and (−2,1). The resulting graph should be a vertical parabola opening upward.
4. To find the equation of a parabola with left and right (LR) points, we use the standard form:
(x-h)² = 4a(y-k) where (h, k) is the vertex of the parabola.
From the given LR points (-2, 4) and (6, 4), we can see that the vertex lies on the line y=4, since the LR points are equidistant from the vertex. Also, since the parabola is opening upward, the value of 'a' must be positive.
Therefore, the vertex is at (h, k) = (2, 4) and a = 1/4a.
Substituting these values in the standard form, we get:
(x-2)² = 4(1/4)(y-4)
(x-2)² = (y-4)
x² - 4x + 4 = y
y = x² - 4x + 4
To sketch the graph, we can plot the vertex at (2, 4) and use the LR points to determine the shape of the parabola. Since the LR points are equidistant from the vertex, the parabola must be symmetrical about the line x=2.
The parts of the parabola are:
Vertex: (2, 4)
Vertex: (2, 4)Axis of symmetry: x = 2
Vertex: (2, 4)Axis of symmetry: x = 2Focus: (2, 15/4)
Vertex: (2, 4)Axis of symmetry: x = 2Focus: (2, 15/4)Directrix: y = 11/4
5. Now, let's find the general equation of a parabola with vertex at (-4, -3), a vertical x-axis, and passing through point (0, -1).
Since the vertex is at (-4, -3), the standard form will be:
(x+4)² = 4a(y+3)
Since the parabola has a vertical axis, we have to use the form (x-h)² = 4a(y-k).
Now, we need to find the value of 'a' to complete the equation. To do this, we use the point (0, -1) that the parabola passes through:
(0+4)² = 4a(-1+3)
16 = 8a
a = 2
Substituting this value of 'a' in the standard form, we get:
(x+4)² = 8(y+3)
Expanding the left-hand side, we get:
x² + 8x + 16 = 8(y+3)
x² + 8x + 16 = 8y + 24
8y = x² + 8x - 8
Dividing both sides by 8, we get the general equation of the parabola:
y = (1/8)x² + x - 1
The parts of the parabola are:
Vertex: (-4, -3)
Vertex: (-4, -3)Axis of symmetry: x = -4
Vertex: (-4, -3)Axis of symmetry: x = -4Focus: (-4, -19/8)
Vertex: (-4, -3)Axis of symmetry: x = -4Focus: (-4, -19/8)Directrix: y = -5/8
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Answer:
1. Find the general equation of a Parabola with Vertex at (2,1) and directrix at x=5.
Since the directrix is a vertical line, the parabola opens horizontally. The focus is located at a distance of a = 3 units to the left of the vertex.
The general equation of a parabola with vertex (h,k), focus (h-a,k), and directrix x = h+a is:
(x - h)^2 = 4a(y - k)
Substituting the given values, we have:
(x - 2)^2 = 12(y - 1)
This is the general form of the parabolic equation. To sketch the graph, we can plot the vertex (2,1) and the point on the directrix that is closest to the vertex, which is (5,1). The focus is located 3 units to the left of the vertex, so it is at (-1,1). We can also plot a few more points on the parabola by substituting values of x and solving for y. The resulting graph should be a horizontal parabola opening to the left.
2. Find the general equation of a Parabola with Focus at (-1, -6) and directrix at y=0.
Since the directrix is a horizontal line, the parabola opens vertically. The vertex is located at the midpoint between the focus and the directrix, which is (−1,−3).
The general equation of a parabola with vertex (h,k), focus (h,k+a), and directrix y = k-a is:
(y - k)^2 = 4a(x - h)
Substituting the given values, we have:
(y + 3)^2 = 24(x + 1)
This is the general form of the parabolic equation. To sketch the graph, we can plot the vertex (−1,−3) and the focus (−1,−6). We can also plot a few more points on the parabola by substituting values of y and solving for x. The resulting graph should be a vertical parabola opening downward.
3. Find the general equation of a Parabola with Vertex at (2,3) and with the end points of Latus Rectum are (6,1) & (-2,1).
The Latus Rectum (LR) is a line segment that passes through the focus and is perpendicular to the axis of symmetry. The length of the LR is equal to 4a, where a is the distance between the vertex and the focus. Since the LR passes through (6,1) and (−2,1), its length is 8 units. Thus, a = 2 units.
The vertex is located at (2,3), which is the midpoint of the LR. Therefore, the focus is located at a distance of 2 units above and below the midpoint, which are (2,5) and (2,1), respectively.
The general equation of a parabola with vertex (h,k), focus (h,k+a), and directrix y = k-a is:
(y - k)^2 = 4a(x - h)
Substituting the given values, we have:
(y - 3)^2 = 16(x - 2)
This is the general form of the parabolic equation. To sketch the graph, we can plot the vertex (2,3) and the focus (2,5) and (2,1). We can also plot the LR by connecting the endpoints (6,1) and (−2,1). The resulting graph should be a vertical parabola opening upward.
4. To find the equation of a parabola with left and right (LR) points, we use the standard form:
(x-h)² = 4a(y-k) where (h, k) is the vertex of the parabola.
From the given LR points (-2, 4) and (6, 4), we can see that the vertex lies on the line y=4, since the LR points are equidistant from the vertex. Also, since the parabola is opening upward, the value of 'a' must be positive.
Therefore, the vertex is at (h, k) = (2, 4) and a = 1/4a.
Substituting these values in the standard form, we get:
(x-2)² = 4(1/4)(y-4)
(x-2)² = (y-4)
x² - 4x + 4 = y
y = x² - 4x + 4
To sketch the graph, we can plot the vertex at (2, 4) and use the LR points to determine the shape of the parabola. Since the LR points are equidistant from the vertex, the parabola must be symmetrical about the line x=2.
The parts of the parabola are:
Vertex: (2, 4)
Vertex: (2, 4)Axis of symmetry: x = 2
Vertex: (2, 4)Axis of symmetry: x = 2Focus: (2, 15/4)
Vertex: (2, 4)Axis of symmetry: x = 2Focus: (2, 15/4)Directrix: y = 11/4
5. Now, let's find the general equation of a parabola with vertex at (-4, -3), a vertical x-axis, and passing through point (0, -1).
Since the vertex is at (-4, -3), the standard form will be:
(x+4)² = 4a(y+3)
Since the parabola has a vertical axis, we have to use the form (x-h)² = 4a(y-k).
Now, we need to find the value of 'a' to complete the equation. To do this, we use the point (0, -1) that the parabola passes through:
(0+4)² = 4a(-1+3)
16 = 8a
a = 2
Substituting this value of 'a' in the standard form, we get:
(x+4)² = 8(y+3)
Expanding the left-hand side, we get:
x² + 8x + 16 = 8(y+3)
x² + 8x + 16 = 8y + 24
8y = x² + 8x - 8
Dividing both sides by 8, we get the general equation of the parabola:
y = (1/8)x² + x - 1
The parts of the parabola are:
Vertex: (-4, -3)
Vertex: (-4, -3)Axis of symmetry: x = -4
Vertex: (-4, -3)Axis of symmetry: x = -4Focus: (-4, -19/8)
Vertex: (-4, -3)Axis of symmetry: x = -4Focus: (-4, -19/8)Directrix: y = -5/8