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2 years ago

Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2.

Mathematics

2 answers:

Anit [1.1K]2 years ago

7 0

Answer:

$y=-\frac{1}{8}x^2$

Step-by-step explanation:

Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2

The distance between the focust and the directrix is the value of 2p

Distance beween focus (0,-2) and y=2 is 4

$2p=4, p=2$

The distance between vertex and focus is p that is 2

Focus is at (0,-2) , so the vertex is at (0,0)

General form of equation is

$y-k=-\frac{1}{4p}(x-h)^2$

where (h,k) is the vertex

Vertex is (0,0) and p = 2

The equation becomes

$y-0=-\frac{1}{4(2)}(x-0)^2$

$y=-\frac{1}{8}x^2$

anyanavicka [17]2 years ago

5 0

Hello,
The parabola having like focus (0,p/2) and as directrix y=-p/2 has as equation x²=2py

Here -p/2=2==>p=-4

x²=-8y is the equation.

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Answer:

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Step-by-step explanation:

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3 years ago

PLEASE HELP NEED THIS ANSWERED CORRECTLY

Helen [10]

f(3) = 1.9

Solution:

Given function:

$$f(x)=\frac{14}{7+2 e^{-0.6 x}}$

To find f(3):

Substitute x = 3 in the given function.

$$f(3)=\frac{14}{7+2 e^{-0.6 (3)}}$

$$f(3)=\frac{14}{7+2 e^{-1.8}}$

Let us first simplify $2e^{-1.8}$ .

Apply exponent rule: $a^{-b}=\frac{1}{a^{b}}$

$2e^{-1.8}=2\frac{1}{e^{1.8}}$

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$$\frac{2}{e^{1.8}}=\frac{2}{6.04964}=0.33059$

$2e^{-1.8}=0.33059$

Substitute this value in f(3).

$$f(3)=\frac{14}{7+0.33059}$

$$f(3)=\frac{14}{7.33059}$

$$f(3)=1.90980$

f(3) = 1.9

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