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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.
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 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)
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 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.
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.
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)
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.
How would you find the vertex p value focus directrix and focal width of negative one fourth x squared equals y?
-(1/4) x2 = y . . . putting this in the standard form x2 = 4cy it becomes : x2 = 4*(-1)y = -4y. This tells us that the parabola is a downward opening parabola with its vertex at the origin(0.0). The focus is at a distance of -1 from the vertex, that is (0,-1). The directrix is equidistant to the focus but on the opposite side of the vertex and is thus the line y = 1. The length of the chord passing through the focus and perpendicular to the major axis is called the Latus Rectum and has a length of 4c. As c = -1 then the length is 4 but again shows as a negative value as it is "below" the vertex.
A parabola is a set of all points that?
are the same distance from a point (known as its focus) and a line (known as its directrix)are given by y=x2, where x is realThere are other characterisations.Apex Answer: are the same distance from a point and a line.
