Wiki User
∙ 11y agoRestate 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.
Wiki User
∙ 11y agoIt is the parabola such that the coordinates of each point on it satisfies the given equation.
No you can't. There is no unique solution for 'x' and 'y'. The equation describes a parabola, and every point on the parabola satisfies the equation.
An equation of a parabola in the x-y plane, is one possibility.
At: x = 6
The equation y = 4x^2 + 5 is a parabola
It is the equation of a parabola.
It is the parabola such that the coordinates of each point on it satisfies the given equation.
No you can't. There is no unique solution for 'x' and 'y'. The equation describes a parabola, and every point on the parabola satisfies the equation.
X equals 0.5at squared is a quadratic equation. It describes a parabola. Y equals mx plus b is a linear equation. It describes a line. You cannot describe a parabola with a linear equation.
When x = -5
An equation of a parabola in the x-y plane, is one possibility.
At: x = 6
y=b+x+x^2 This is a quadratic equation. The graph is a parabola. The quadratic equation formula or factoring can be used to solve this.
Line of symmetry: x = 3
The equation y = 4x^2 + 5 is a parabola
The given equation is not that of a parabola.
A quadratic equation (t=s2+3). This kind of line will result in a parabola like curve.