Question

What is the equation of the quadratic graph with a focus of (1, 1) and a directrix of y = −1? PLs HELP!!

Answer 1

ANSWER
$f(x)= \frac{1}{4} {(x - 1)}^{2}$

EXPLANATION

Since the directrix is
$y = - 1$

the axis of symmetry of the parabola is parallel to the y-axis.

Again, the focus being,

$(1,1)$

also means that the parabola will open upwards.

The equation of parabola with such properties is given by,

${(x - h)}^{2} = 4p(y - k)$
where
$(h,k)$

is the vertex of the parabola.

The directrix and the axis of symmetry of the parabola will intersect at
$(1, - 1)$

The vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix.

This implies that,
$h = \frac{1 + 1}{2} = 1$

and
$k = \frac{ - 1 + 1}{2} = 0$

The equation of the parabola now becomes,

$(x - 1) ^{2} = 4p(y - 0)$

$|p| = 1$

Thus, the distance between the vertex and the directrix.

This means that,

$p = - 1 \: or \: 1$

Since the parabola opens up, we choose
$p = 1$
Our equation now becomes,

${(x - 1)}^{2} = 4(1)(y - 0)$

This simplifies to
${(x - 1)}^{2} = 4y$

or

$y = \frac{1}{4} {(x - 1)}^{2}$

This is the same as,

$f(x)= \frac{1}{4} {(x - 1)}^{2}$

The correct answer is D .