How the equation of directrix is y plus a equals 0 in parabola?

Answer 1

Restate the question: "In the parabola y = ax2, why is the equation of the directrix y+a = 0?

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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.