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The focus of a parabola is a specific point that defines its shape, while the directrix is a line used in the definition of a parabola. If the directrix is given as ( y = -2 ), the parabola opens either upwards or downwards. The focus would be located at a point above or below this directrix, depending on the orientation of the parabola. Specifically, for a parabola that opens upwards, the focus would be positioned at ( (h, k + p) ), where ( p ) is the distance from the vertex to the focus, and the vertex would be located at ( (h, -2 + p) ).
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What is the equation of the directrix of the parabola?
The answer depends on the form in which the equation of the parabola is given. For y^2 = 4ax the directrix is x = -2a.
How do you get directrix of a parabola?
The directrix of a parabola can be found using its standard form equation. For a parabola that opens upwards or downwards, given by (y = ax^2 + bx + c), the directrix is located at (y = k - \frac{1}{4p}), where (k) is the vertex's y-coordinate and (p) is the distance from the vertex to the focus. For a parabola that opens sideways, the directrix is given by (x = h - \frac{1}{4p}), where (h) is the vertex's x-coordinate. The value of (p) can be determined based on the coefficients of the quadratic equation.
What is the equation of a parabola with vertex (0 0) and directrix x -3.?
The equation of a parabola with vertex at (0, 0) and a directrix of ( x = -3 ) opens to the right, as the directrix is a vertical line. The distance from the vertex to the directrix is 3 units. The standard form of the equation for a horizontally-opening parabola is given by ( y^2 = 4px ), where ( p ) is the distance from the vertex to the directrix. Therefore, with ( p = 3 ), the equation is ( y^2 = 12x ).
What is the focus and directrix of y equals 8x squared?
Need set of point parabola passes through.X = 1Y = 8----------------(1, 8)For focus,X2 = 4pYinsert values, solve for p(1)2 = 4p(8)= 1/32=======focus- 1/32=======directrix( long time since I have done this. Formula is correct, check my math )
What is the standard form of the equation of the parabola with vertex 00 and directrix y4?
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
What is the equation of the quadratic graph with a focus of (3 6) and a directrix of y 4?
For a parabola with a y=... directrix, it is of the form: (x - h)^2 = 4p(y - k) with vertex (h, k), focus (h, k + p) and directrix y = k - p With a focus of (3, 6) and a directrix of y = 4, this means: (h, k + p) = (3, 6) → k + p = 6 y = k - p = 4 → k = 5, p = 1 (solving the simultaneous equations) → vertex is (3, 5) → parabola is (x - 3)^2 = 4(y - 5) which can be rearranged into y = 1/4 x^2 - 3/2 x + 29/4
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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What is the relationship between the focus and directrix?
i assume this is locus you are talking about, in which case: they are both the same distance from the vertex - focal length, focus is a point: P(x,y) and directrix is a horizontal line e.g. y=-1
What is the equation of the directrix of the parabola?
The answer depends on the form in which the equation of the parabola is given. For y^2 = 4ax the directrix is x = -2a.
How do you get directrix of a parabola?
The directrix of a parabola can be found using its standard form equation. For a parabola that opens upwards or downwards, given by (y = ax^2 + bx + c), the directrix is located at (y = k - \frac{1}{4p}), where (k) is the vertex's y-coordinate and (p) is the distance from the vertex to the focus. For a parabola that opens sideways, the directrix is given by (x = h - \frac{1}{4p}), where (h) is the vertex's x-coordinate. The value of (p) can be determined based on the coefficients of the quadratic equation.
What is the equation of a parabola with vertex (0 0) and directrix x -3.?
The equation of a parabola with vertex at (0, 0) and a directrix of ( x = -3 ) opens to the right, as the directrix is a vertical line. The distance from the vertex to the directrix is 3 units. The standard form of the equation for a horizontally-opening parabola is given by ( y^2 = 4px ), where ( p ) is the distance from the vertex to the directrix. Therefore, with ( p = 3 ), the equation is ( y^2 = 12x ).
What is the focus and directrix of y equals 8x squared?
Need set of point parabola passes through.X = 1Y = 8----------------(1, 8)For focus,X2 = 4pYinsert values, solve for p(1)2 = 4p(8)= 1/32=======focus- 1/32=======directrix( long time since I have done this. Formula is correct, check my math )
What is the standard form of the equation of the parabola with vertex 00 and directrix y4?
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
What is the axis of symmetry for the parabola with vertex (-2 -4) and directrix y 1?
The axis of symmetry is x = -2.
How the equation of directrix is y plus a equals 0 in parabola?
Restate the question: "In the parabola y = ax2, why is the equation of the directrix y+a = 0?If this is not your question, please clarify and ask the question again.The "locus" definition of a parabola says that a parabola is the set of all points which are the same distance from a given point and a given line. The point is called the focus, F. The line is called the directrix, d.With a little foresight, we set things up so that the vertex of the parabola is at the origin O, and the parabola opens upward: Let the equation of d be y = -a ... which can be written y+a+0 ... , and let F be (0,a). If you make a sketch, it is clear that the distance from O to F is a, and the shortest (perpendicular) distance from O to d is also a. This shows that the origin is on the parabola.To set up an equation, we need formulas for the distance from a point P(x,y) on the parabola to F and to d:Mark a point in the first quadrant and label it P(x,y). The distance from P to the x-axis is y, and the distance from the x-axis to d is a. The distance from P to d is y+a.To find the distance from P to F, we use the distance formula:PF = sqrt((x2-x1)2+(y2-y1)2) = sqrt((x-0)2+(y-a)2) = sqrt(x2+(y-a)2).If P is on the parabola then PF = Pd sqrt(x2+(y-a)2) = y+a.Square both sides to get x2+(y-a)2 = (y+a)2 x2+y2-2ay+a2 = y2+2ay+a2 x2-2ay = 2ay x2 = 4ay 4ay = x2 y = (1/(4a))x2.So, y+a=0 isn't the directrix of y=ax2 after all, it's actually the directix of y=(1/(4a))x2.Common practice is to replace a by p and switch the equation around:y=(1/(4p))x2 4py=x2 x2=4py. This is the equation of a parabola with focus (0,p) and directrix y=-p.
What is the focus of a parabola?
The focus of a parabola is a fixed point that lies on the axis of the parabola "p" units from the vertex. It can be found by the parabola equations in standard form: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h) depending on the shape of the parabola. The vertex is defined by (h,k). Solve for p and count that many units from the vertex in the direction away from the directrix. (your focus should be inside the curve of your parabola)
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)
