1. To sketch the graph of y^2-40x=0, we can first rearrange it to the standard form:
(y-0)^2 = 4(10)x
This is a rightward opening parabola with vertex at (0,0) and a horizontal axis of symmetry. The value of 4a is 40, so a=10.
The vertex form of this equation is (y-0)^2 = 4(10)(x-0), which gives us h=0 and k=0. Therefore, the vertex is at the origin.
To sketch the graph, we can plot the vertex at (0,0), and then find the x-intercepts by setting y=0:
0^2 - 40x = 0
x = 0
So the graph passes through the point (0,0) and the x-axis at x=0. To find the focus and directrix, we can use the formula:
4a = 40
a = 10
The focus is located at (a,0) = (10,0), and the directrix is the line x=-a = -10.
So the graph of y^2-40x=0 is a rightward opening parabola with vertex at the origin, passing through (0,0) and the x-axis at x=0, with focus at (10,0) and directrix at x=-10.
2. Identify the general form of a Parabola if the vertex is at (0,0) and the directrix is at y=-3.
Since the vertex is at (0,0), we know that the standard form of the parabola is either (x-h)^2 = 4a(y-k) or (y-k)^2 = 4a(x-h). Since the directrix is a horizontal line, we know that the parabola opens upward or downward, so the standard form must be (y-k)^2 = 4a(x-h).
The vertex is at (0,0), so h = 0 and k = 0. The directrix is y = -3, which means that the distance from any point on the parabola to the directrix is 3 units. Since the parabola opens either upward or downward, we know that a is either positive or negative, respectively.
Using the formula for the distance from a point (x,y) to a horizontal line y = c, we get:
|y - (-3)| = 3
|y + 3| = 3
Squaring both sides, we get:
(y + 3)^2 = 9 or (y - 3)^2 = 9
Expanding, we get:
y^2 + 6y + 9 = 9 or y^2 - 6y + 9 = 9
Simplifying, we get:
y^2 + 6y = 0 or y^2 - 6y = 0
Factoring out y, we get:
y(y + 6) = 0 or y(y - 6) = 0
So the two possible equations of the parabola are:
(y - 3)^2 = 4ax or (y + 3)^2 = 4ax
I
f we plug in the vertex (0,0), we get:
(0 - 3)^2 = 4a(0) or (0 + 3)^2 = 4a(0)
9 = 0 or 9 = 0
So the only possible equation is:
(y + 3)^2 = 4ax
T
his is the standard form of the parabola. To find the general form, we can expand it:
y^2 + 6y + 9 = 4ax
4a = coefficient of x
So the general form is:
4ax - y^2 - 6y - 9 = 0.
3. Since the directrix is a vertical line, the parabola must open horizontally. The focus is located to the right of the directrix, so the parabola must open to the right. Therefore, the axis of symmetry is a vertical line passing through the focus and the vertex.
To find the equation of the parabola, we need to determine the distance between the focus and the directrix, which is equal to the distance between the vertex and the focus. This distance is given by the absolute value of the parabola's parameter, denoted as "a" in standard form. In this case, a = 4.
The vertex lies on the axis of symmetry and is equidistant from the focus and directrix. Since the directrix is the vertical line x = -4, the vertex must be located at the point (h, k) = (0, 0).
Using the standard form for a horizontally opening parabola, we can write:
(x - h)² = 4a (y - k)
Substituting the values of h, k, and a, we get:
(x - 0)² = 4(4) (y - 0)
Simplifying and rearranging, we obtain the general form of the equation:
x² - 16y = 0
This is the equation of a parabola with vertex at (0,0), opening to the right, and with focus at (4,0) and directrix x = -4.
4. Since the axis of symmetry is y=0 and the vertex is (0,0), the parabola opens either upward or downward. Additionally, we have 4a=-16, which means a=-4.
Using the standard form for a parabola that opens upward or downward with vertex at (h,k), we have:
(x-h)^2 = 4a(y-k)
Plugging in the given values, we get:
x^2 = -16y
To put this equation into general form, we can move all the terms to one side:
x^2 + 0x + 16y + 0 = 0
Therefore, the general form of the parabola is x^2 + 16y = 0.
Answers & Comments
Verified answer
Answer:
1. To sketch the graph of y^2-40x=0, we can first rearrange it to the standard form:
(y-0)^2 = 4(10)x
This is a rightward opening parabola with vertex at (0,0) and a horizontal axis of symmetry. The value of 4a is 40, so a=10.
The vertex form of this equation is (y-0)^2 = 4(10)(x-0), which gives us h=0 and k=0. Therefore, the vertex is at the origin.
To sketch the graph, we can plot the vertex at (0,0), and then find the x-intercepts by setting y=0:
0^2 - 40x = 0
x = 0
So the graph passes through the point (0,0) and the x-axis at x=0. To find the focus and directrix, we can use the formula:
4a = 40
a = 10
The focus is located at (a,0) = (10,0), and the directrix is the line x=-a = -10.
So the graph of y^2-40x=0 is a rightward opening parabola with vertex at the origin, passing through (0,0) and the x-axis at x=0, with focus at (10,0) and directrix at x=-10.
2. Identify the general form of a Parabola if the vertex is at (0,0) and the directrix is at y=-3.
Since the vertex is at (0,0), we know that the standard form of the parabola is either (x-h)^2 = 4a(y-k) or (y-k)^2 = 4a(x-h). Since the directrix is a horizontal line, we know that the parabola opens upward or downward, so the standard form must be (y-k)^2 = 4a(x-h).
The vertex is at (0,0), so h = 0 and k = 0. The directrix is y = -3, which means that the distance from any point on the parabola to the directrix is 3 units. Since the parabola opens either upward or downward, we know that a is either positive or negative, respectively.
Using the formula for the distance from a point (x,y) to a horizontal line y = c, we get:
|y - (-3)| = 3
|y + 3| = 3
Squaring both sides, we get:
(y + 3)^2 = 9 or (y - 3)^2 = 9
Expanding, we get:
y^2 + 6y + 9 = 9 or y^2 - 6y + 9 = 9
Simplifying, we get:
y^2 + 6y = 0 or y^2 - 6y = 0
Factoring out y, we get:
y(y + 6) = 0 or y(y - 6) = 0
So the two possible equations of the parabola are:
(y - 3)^2 = 4ax or (y + 3)^2 = 4ax
I
f we plug in the vertex (0,0), we get:
(0 - 3)^2 = 4a(0) or (0 + 3)^2 = 4a(0)
9 = 0 or 9 = 0
So the only possible equation is:
(y + 3)^2 = 4ax
T
his is the standard form of the parabola. To find the general form, we can expand it:
y^2 + 6y + 9 = 4ax
4a = coefficient of x
So the general form is:
4ax - y^2 - 6y - 9 = 0.
3. Since the directrix is a vertical line, the parabola must open horizontally. The focus is located to the right of the directrix, so the parabola must open to the right. Therefore, the axis of symmetry is a vertical line passing through the focus and the vertex.
To find the equation of the parabola, we need to determine the distance between the focus and the directrix, which is equal to the distance between the vertex and the focus. This distance is given by the absolute value of the parabola's parameter, denoted as "a" in standard form. In this case, a = 4.
The vertex lies on the axis of symmetry and is equidistant from the focus and directrix. Since the directrix is the vertical line x = -4, the vertex must be located at the point (h, k) = (0, 0).
Using the standard form for a horizontally opening parabola, we can write:
(x - h)² = 4a (y - k)
Substituting the values of h, k, and a, we get:
(x - 0)² = 4(4) (y - 0)
Simplifying and rearranging, we obtain the general form of the equation:
x² - 16y = 0
This is the equation of a parabola with vertex at (0,0), opening to the right, and with focus at (4,0) and directrix x = -4.
4. Since the axis of symmetry is y=0 and the vertex is (0,0), the parabola opens either upward or downward. Additionally, we have 4a=-16, which means a=-4.
Using the standard form for a parabola that opens upward or downward with vertex at (h,k), we have:
(x-h)^2 = 4a(y-k)
Plugging in the given values, we get:
x^2 = -16y
To put this equation into general form, we can move all the terms to one side:
x^2 + 0x + 16y + 0 = 0
Therefore, the general form of the parabola is x^2 + 16y = 0.