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)= (1)/(4) {(x - 1)}^(2)`

Step-by-step 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 = (1 + 1)/(2) = 1`

and

`k = ( - 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 = (1)/(4) {(x - 1)}^(2)`

This is the same as,


`f(x)= (1)/(4) {(x - 1)}^(2)`

The correct answer is D .