Question

What is the equation of the quadratic graph with a focus of (3, 6) and a directrix of y=4

Answer 1

Vertex is at

(

3

,

−

1

)

, focus is at

(

3

,

−

15

16

)

and

directrix is

y

=

−

1

1

16

.

Explanation:

y

=

4

(

x

−

3

)

2

−

1

Comparing with standard form of vertex form equation

y

=

a

(

x

−

h

)

2

+

k

;

(

h

,

k

)

being vertex , we find here

h

=

3

,

k

=

−

1

,

a

=

4

. So vertex is at

(

3

,

−

1

)

.

Vertex is at equidistance from focus and directrix and at opposite

sides . The distance of vertex from directrix is

d

=

1

4

|

a

|

∴

d

=

1

4

⋅

4

=

1

16

. since

a

>

0

, the parabola opens upwards and

directrix is below vertex. So directrix is

y

=

(

−

1

−

1

16

)

=

−

17

16

=

−

1

1

16

and focus is at

(

3

,

(

−

1

+

1

16

)

)

or

(

3

,

−

15

16

)

Explanation:

Answer 2

Check the picture below.

one thing to notice is that, the directrix is the horizontal line y = 4, meaning is a vertical parabola, and thus the squared variable is the "x". The directrix is below the focus point, meaning the parabola is opening upwards, it also means that the "p" distance is positive, in this case p = 1.

The vertex is always half-way between the focus point and the directrix, in this case is at (3,5).

`\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{we'll use this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\bf \begin{cases} h=3\\ k=5\\ p=1 \end{cases}\implies 4(1)(y-5)=(x-3)^2\implies y-5=\cfrac{1}{4}(x-3)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=\cfrac{1}{4}(x-3)^2+5~\hfill`