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

Answers: 2

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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

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

Answer from: Quest

c 1 hour

step-by-step explanation:

m = 2e

m = 2 so 2 = 2e

1 = e

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