How to derive the equation of a parabola from the directrix and focus

Question

Could someone please offer me proof and explanation of the following? - I am just having trouble with finding the '$a$' part of the equation.

"The leading coefficient '$a$' in the equation

$$y−y_1 =​​ a(x−x_1)^{2}$$​ indicates how "wide" and in what direction the parabola opens. It's always the reciprocal of $2$ times the distance from the directrix to the focus."

Thanks.

Answer 1

Your equation $y-y_1=a(x-x_1)^2$ comes from a horizontal directrix and a focus not on the directrix.

So given those requirements, let the focus be the point $(x_1,y_1+d/2)$ and the directrix be the line $y=y_1-d/2$, where $d\ne 0$. This will put the vertex of the parabola at $(x_1,y_1)$ and the distance between the focus and the directrix will be $|d|$. Then the parabola is the locus of points equidistant from the focus and the directrix. In other words, the distance of point $(x,y)$ to point $(x_1,y_1+d/2)$ equals the distance of point $(x,y)$ to the line $y=y_1-d/2$. Writing this as an equation,

$$\sqrt{(x-x_1)^2+(y-(y_1+d/2))^2}=|y-(y_1-d/2)|$$

Squaring each side, expanding the squared terms in $y$, and canceling equal terms gets us

$$(x-x_1)^2-dy+dy_1=dy-dy_1$$

which leads us to

$$y-y_1=\frac 1{2d}(x-x_1)^2$$

which is what you wanted. However, note the one mistake in the quote in your question. $a$ may not be "the reciprocal of 2 times the distance from the directrix to the focus." It may the negative of that reciprocal.

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Here is some help in "expanding the squared terms in $y$". On the left hand side,

$$(y-(y_1+d/2))^2=y^2-2y(y_1+d/2)+(y_1+d/2)^2$$ $$=y^2-2yy_1-dy+y_1^2+dy_1+d^2/4$$

On the right hand side,

$$(y-(y_1-d/2))^2=y^2-2y(y_1-d/2)+(y_1-d/2)^2$$ $$=y^2-2yy_1+dy+y_1^2-dy_1+d^2/4$$

Answer 2

The parabola is defined by the locus of the points that are at equal distance from the focus and a line, called the directrice.

Let's call $d$ the distance between the focus $F$ and the directrix $D$.

Then $a=\dfrac{1}{2d}$