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Which equation represents a parabola opening to the right with a vertex at the origin and a focus at (40)?
The equation that represents a parabola opening to the right with its vertex at the origin (0,0) and a focus at (4,0) is given by ( y^2 = 4px ), where ( p ) is the distance from the vertex to the focus. Since the focus is located at (4,0), ( p = 4 ). Therefore, the equation of the parabola is ( y^2 = 16x ).
What is the equation of the parabola when the vertex is (3 2) and the focus is (5 2)?
The equation of a parabola can be derived from its vertex and focus. Given the vertex at (3, 2) and the focus at (5, 2), the parabola opens to the right. The standard form of the equation is ((y - k)^2 = 4p(x - h)), where ((h, k)) is the vertex and (p) is the distance from the vertex to the focus. Here, (h = 3), (k = 2), and (p = 2) (the distance between x-coordinates of the vertex and focus), leading to the equation ((y - 2)^2 = 4(x - 3)).
How do you write an equation of a parabola with vertex at the origin and the given focus 60?
To write the equation of a parabola with its vertex at the origin (0, 0) and a focus at (0, 60), you first identify the orientation of the parabola. Since the focus is above the vertex, the parabola opens upwards. The standard form of the equation for a parabola that opens upwards is ( y = \frac{1}{4p}x^2 ), where ( p ) is the distance from the vertex to the focus. Here, ( p = 60 ), so the equation becomes ( y = \frac{1}{240}x^2 ).
What is the equation of the parabola with focuse (07) and the directrix y1?
To find the equation of the parabola with focus at (0, 7) and directrix ( y = 1 ), we first determine the vertex, which is the midpoint between the focus and the directrix. The vertex is at ( (0, 4) ). The distance from the vertex to the focus is 3, so the parabola opens upward. The equation of the parabola can be expressed as ( (x - h)^2 = 4p(y - k) ), where ( (h, k) ) is the vertex and ( p ) is the distance from the vertex to the focus. Thus, the equation is ( x^2 = 12(y - 4) ).
What is the equation of the parabola with focus (1 3) and vertex at (3 3)?
The equation of a parabola can be determined using its focus and vertex. Given the focus at (1, 3) and the vertex at (3, 3), the parabola opens horizontally since the x-coordinate of the focus is less than that of the vertex. The standard form for a horizontally opening parabola is ((y - k)^2 = 4p(x - h)), where (h, k) is the vertex and p is the distance from the vertex to the focus. Here, (p = -2) (the focus is 2 units left of the vertex), so the equation is ((y - 3)^2 = -8(x - 3)).
Which equation represents a parabola opening to the right with a vertex at the origin and a focus at (40)?
The equation that represents a parabola opening to the right with its vertex at the origin (0,0) and a focus at (4,0) is given by ( y^2 = 4px ), where ( p ) is the distance from the vertex to the focus. Since the focus is located at (4,0), ( p = 4 ). Therefore, the equation of the parabola is ( y^2 = 16x ).
What is the equation of the parabola when the vertex is (3 2) and the focus is (5 2)?
The equation of a parabola can be derived from its vertex and focus. Given the vertex at (3, 2) and the focus at (5, 2), the parabola opens to the right. The standard form of the equation is ((y - k)^2 = 4p(x - h)), where ((h, k)) is the vertex and (p) is the distance from the vertex to the focus. Here, (h = 3), (k = 2), and (p = 2) (the distance between x-coordinates of the vertex and focus), leading to the equation ((y - 2)^2 = 4(x - 3)).
How do you write an equation of a parabola with vertex at the origin and the given focus 60?
To write the equation of a parabola with its vertex at the origin (0, 0) and a focus at (0, 60), you first identify the orientation of the parabola. Since the focus is above the vertex, the parabola opens upwards. The standard form of the equation for a parabola that opens upwards is ( y = \frac{1}{4p}x^2 ), where ( p ) is the distance from the vertex to the focus. Here, ( p = 60 ), so the equation becomes ( y = \frac{1}{240}x^2 ).
What is the equation of the parabola with focuse (07) and the directrix y1?
To find the equation of the parabola with focus at (0, 7) and directrix ( y = 1 ), we first determine the vertex, which is the midpoint between the focus and the directrix. The vertex is at ( (0, 4) ). The distance from the vertex to the focus is 3, so the parabola opens upward. The equation of the parabola can be expressed as ( (x - h)^2 = 4p(y - k) ), where ( (h, k) ) is the vertex and ( p ) is the distance from the vertex to the focus. Thus, the equation is ( x^2 = 12(y - 4) ).
What is the equation of a parabola with a vertex at 0 0 and a focus at 0 6?
The standard equation for a Parabola with is vertex at the origin (0,0) is, x2 = 4cy if the parabola opens vertically upwards/downwards, or y2 = 4cx when the parabola opens sideways. As the focus is at (0,6) then the focus is vertically above the vertex and we have an upward opening parabola. Note that c is the distance from the vertex to the focus and in this case has a value of 6 (a positive number). The equation is thus, x2 = 4*6y = 24y
What is the equation of the parabola with focus (1 3) and vertex at (3 3)?
The equation of a parabola can be determined using its focus and vertex. Given the focus at (1, 3) and the vertex at (3, 3), the parabola opens horizontally since the x-coordinate of the focus is less than that of the vertex. The standard form for a horizontally opening parabola is ((y - k)^2 = 4p(x - h)), where (h, k) is the vertex and p is the distance from the vertex to the focus. Here, (p = -2) (the focus is 2 units left of the vertex), so the equation is ((y - 3)^2 = -8(x - 3)).
How do you find the equation of the parabola when you have the vertex focus and directrix?
To find the equation of a parabola given the vertex, focus, and directrix, start by identifying the vertex coordinates ((h, k)), the focus ((h, k + p)) for a vertical parabola (or ((h + p, k)) for a horizontal one), and the distance (p) from the vertex to the focus. The directrix will be a line located at (y = k - p) for vertical parabolas or (x = h - p) for horizontal ones. The standard form of the equation is ((x - h)^2 = 4p(y - k)) for vertical parabolas and ((y - k)^2 = 4p(x - h)) for horizontal parabolas. Substitute (p) with the distance calculated from the vertex to the focus or directrix to finalize the equation.
What is the standard equation of a parabola that opens up or down and whose vertex is at the origin?
focus , directrix
What is the standard equation for vertex at origin opens down 1 and 76 units between the vertex and focus?
Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
What is the equation of a parabola with Vertex (0 0) and focus (-3 0) a.y2-4x b.y2-12x c.y24x d.y212x?
The equation of a parabola with a vertex at the origin (0, 0) and a focus at (-3, 0) opens to the left. The standard form for such a parabola is ( y^2 = -4px ), where ( p ) is the distance from the vertex to the focus. Here, ( p = 3 ), so the equation becomes ( y^2 = -12x ). Therefore, the correct answer is b) ( y^2 = -12x ).
What does 4 stands for in equation of parabola square of square of y equals 4ax?
A parabola with an equation, y2 = 4ax has its vertex at the origin and opens to the right. It's not just the '4' that is important, it's '4a' that matters. This type of parabola has a directrix at x = -a, and a focus at (a, 0). By writing the equation as it is, the position of the directrix and focus are readily identifiable. For example, y2 = 2.4x doesn't say a great deal. Re-writing the equation of the parabola as y2 = 4*(0.6)x tells us immediately that the directrix is at x = -0.6 and the focus is at (0.6, 0)
Using the given equations of parabolas find the focus the directrix and the equation of the axis of symmetry of x2 -8y?
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