Find the equation of the parabola with a focus (2, 5) and directrix of y = 3.

A.)

y=x^2/4 + x + 5

B.)

y=x^2/4 + x - 5

C.)

y=x^2/4 - x - 5

D.)

y=x^2/4 - x + 5

One form of an equation is given by

y=a (x-h

)

2

+k

y=a (x-h) 2+k

where

(h, k)

(h, k)

the coordinates of the vertex and

(h, k+

1

4a

)

The parabola is symmetric with respect to

y=x

y=x

and can be viewed as a standard downward parabola with a rotation of 45 degrees clockwise. So, follow the steps below to obtain its equation.

1) The length between the focus and vertex is

f=

4

2

+

4

2

-

-

-

-

-

-

√

=4

2

-

√

f=42+42=42

. The standard equation is

y=-

1

4f

x

2

y=-14fx2

.

2) Shift the vertex to (-2, - 2),

y+2=-

1

16

2

-

√

(x+2

)

2

y+2=-1162 (x+2) 2

3) Rotate the equation - 45-degrees

x→

1

2

√

(x-y)

x→12 (x-y)

,

y→-

1

2

√

(x+y)

y→-12 (x+y)

to get

-

1

2

-

√

(x+y) + 2=-

1

16

2

-

√

(

1

2

-

√

(x-y) + 2)

2