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stepan [7]
3 years ago
14

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

Mathematics
1 answer:
Delicious77 [7]3 years ago
4 0
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 .
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