Question

1. Find the standard form of the equation of the parabola with a focus at (0, -9) and a directrix y = 9.
A ) y = negative one divided by ninex2
B ) y2 = -36x
C ) y = negative one divided by thirty sixx2
D ) y2 = -9x

2. Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, 9).

A ) y = one divided by thirty sixx2
B ) y = one divided by ninex2
C ) y2 = 9x
D ) y2 = 36x

3. Find the vertex, focus, directrix, and focal width of the parabola.

x = 4y2

Vertex: (0, 0); Focus: one divided by four comma zero; Directrix: x = negative one divided by four; Focal width: 0.25
Vertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = one divided by sixteen; Focal width: 16
Vertex: (0, 0); Focus: zero comma one divided by sixteen; Directrix: y = negative one divided by sixteen; Focal width: 16
Vertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = negative one divided by sixteen; Focal width: 0.25

4. Find the standard form of the equation of the parabola with a focus at (3, 0) and a directrix at x = -3.

A ) y = one divided by twelvex2
B ) -12y = x2
C ) x = one divided by twelvey2
D ) y2 = 6x

Answer 1

Answer:

Part 1) Option C)  y = negative one divided by thirty six x²

Part 2) Option A) y = one divided by thirty six x²

Part 3) Vertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = negative one divided by sixteen; Focal width: 0.25

Part 4) Option  C) x = one divided by twelve y²

Step-by-step explanation:

Part 1) Find the standard form of the equation of the parabola with a focus at (0, -9) and a directrix y = 9.

we know that

The vertex form of the equation of the vertical parabola is equal to

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

where

Vertex ----> (h,k)

Focus ----> F(h,k+p)

directrix ----->  y=k-p

we have

F(0,-9)

so

h=0

k+p=-9 -----> equation A

y=9

so

k-p=9 ----> equation B

Adds equation A and equation B

k+p=-9

k-p=9

-----------

2k=0

k=0

so

Find the value of p

0+p=-9

p=-9

substitute in the equation

$(x-0)^{2}=4(-9)(y-0)$

$(x)^{2}=-36y$

Convert to standard form

isolate the variable y

$y=-\frac{1}{36}x^{2}$

Part 2) Find the standard form of the equation of the parabola with a vertex at the origin and a focus  at (0, 9)

we know that

The vertex form of the equation of the vertical parabola is equal to

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

where

Vertex ----> (h,k)

Focus ----> F(h,k+p)

directrix ----->  y=k-p

we have

Vertex (0,0) -----> h=0,k=0

F(0,9)

so

k+p=9------> 0+p=9 -----> p=9

substitute in the equation

$(x-0)^{2}=4(9)(y-0)$

$(x)^{2}=36(y)$

Convert to standard form

isolate the variable y

$y=\frac{1}{36}x^{2}$

Part 3)  Find the vertex, focus, directrix, and focal width of the parabola.  

x = 4y²

we know that

The vertex form of the equation of the horizontal parabola is equal to

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

where

Vertex ----> (h,k)

Focus ----> F(h+p,k)

directrix ----->  x=h-p

we have

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

so

Vertex (0,0) ------> h=0,k=0

4p=1/4 ------> Focal width

p=1/16

Focus F(0+1/16,0) ----> F(1/16,0)

directrix -----> x=0-1/16 -----> x=-1/16

therefore

Vertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = negative one divided by sixteen; Focal width: 0.25

Part 4)  Find the standard form of the equation of the parabola with a focus at (3, 0) and a directrix at x = -3.

we know that

The vertex form of the equation of the horizontal parabola is equal to

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

where

Vertex ----> (h,k)

Focus ----> F(h+p,k)

directrix ----->  x=h-p

we have

Focus F(3,0)

so

h+p=3 ------> equation A

k=0

directrix x=-3

so

h-p=-3 ------> equation B

Adds equation A and equation B and solve for h

h+p=3

h-p=-3

------------

2h=0 -----> h=0

Find the value of p

h+p=3 ------> 0+p=3 ------> p=3

substitute in the equation

$(y-0)^{2}=4(3)(x-0)$

$(y)^{2}=12(x-0)$

Convert to standard form

isolate the variable x

$x=\frac{1}{12}y^{2}$