Mathematics, 02.08.2019 21:30 vianney84

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

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Answer from: 1937

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)

h=0

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

y=9

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

Adds equation A and equation B

k+p=-9

k-p=9

-----------

2k=0

k=0

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)

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

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

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)

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

k=0

directrix x=-3

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

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

Answer from: Quest

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step-by-step explanation:

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Answer from: Quest

correct

step-by-step explanation:

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