Question

Mark the focus of the parabola you are going to create at F(6, 4). Draw a horizontal line that is 6 units below the focus. This line will be the directrix of your parabola. What is the equation of the line?

Answer 1

Answer:

The equation of the line is y = -2,

Step-by-step explanation:

In the coordinate plane (x, y) represents the position of a point,

Where, x shows the location towards x-axis,

While, y shows the location towards y-axis,

Given,

The focus of the parabola is F(6, 4),

That is, the focus will be located 6 unit right from the (0,0),

And, 4 unit up from (6, 0),

Let the equation of directrix,

y = mx + c

Also, directrix is 6 unit below the focus,

Thus the coordinates of a point on the line are (6, -2),

Also, the line is horizontal,

⇒ m = 0 ( slope about x -axis )

$-2 = 0(6) + c\implies c = -2$

Hence, the equation of directrix,

$y=-2$

Answer 2

Answer:

(You don't need a picture to find the equation.  It's all algebra)

The equation is, depending upon the form you're looking for:

$(x-6)^2=12(y-1)$ or

$y=\frac{1}{12}x^2-x+4$

Step-by-step explanation:

When you plot the focus and draw in the directrix, you see that the focus is above the directrix.  That means that, since a parabola "hugs" the focus, this parabola opens upwards and is a "y = ax^2" type.  By nature, the vertex is exactly halfway between the focus and the directrix on the same x-coordinate as the focus.  So the coordinates of the vertex, (h, k), are (6, 1).  The p value is the distance between the vertex and either the focus or the directrix (they're the same distance) so p = 3.  The standard form of the parabola we are looking for is:

$(x-h)^2=4p(y-k)$

Filling in, we have, before simplification:

$(x-6)^2=4(3)(y-1)$ and

$(x-6)^2=12(y-1)$

If we expand the left side and collect like terms we will get the standard $y=ax^2+bx+c$ version, which is

$y=\frac{1}{12}x^2-x+4$

It depends upon whether you need one form or the other.  They're the same equation in different forms.