Drag the focus and directrix to define a parabola with the equation:
\qquad y - PRETTY_Y1 = PRETTY_A(x - PRETTY_X1)^2
.
Equation of the parabola:
y
{}+0
{}={}
\dfrac{1}{4}
(x
{}+0
)^2
The equation is in the form y - y_1 = a(x - x_1)^2
.
The leading coefficient, a
, in the equation is PRETTY_A
, which is positive.
Therefore the parabola is upward opening.
The leading coefficient, a
, in the equation is PRETTY_A
, which is negative.
Therefore the parabola is downward opening.
The value of x_1
is PRETTY_X1
, what does this number mean for the parabola?
The minimum value of PRETTY_A(x - PRETTY_X1)^2
occurs when
(x - PRETTY_X1)^2 = 0
.
Therefore the vertex of the parabola is at x = PRETTY_X1
.
The maximum value of PRETTY_A(x - PRETTY_X1)^2
occurs when
(x - PRETTY_X1)^2 = 0
.
Therefore the vertex of the parabola is at x = PRETTY_X1
.
x
-coordinate as the vertex,
so move the focus horizontally, so its x
-coordinate is PRETTY_X1
.
At x = PRETTY_X1
, then y - PRETTY_Y1 = 0
.
Therefore, the vertex of the parabola is at y = PRETTY_Y1
.
The focus and directrix should be an equal distance above and below this value.
The number in the denominator of a
is twice the distance between the directrix to the focus.
Therefore, the focus and directrix are separated by 1 / (2 * abs(A))
units.
The focus is therefore 1 / (4 * A)
units above the vertex,
and the directrix is 1 / (4 * A)
units below the vertex.
The focus is therefore 1 / (4 * -A)
units below the vertex,
and the directrix is 1 / (4 * -A)
units above the vertex.
(PRETTY_X1, PRETTY_FOCUS_Y)
and the directrix is at y = PRETTY_DIR_Y
.