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Determine the equation of the parabola that opens up, has focus
(0,4), and a focal diameter of 28 .

Determine the equation of the parabola that opens up, has focus $(0,4)$ , and a focal diameter of $28$ .

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Write equations of parabolas in vertex form using the focus and directrix

Full solution

Q. Determine the equation of the parabola that opens up, has focus

(0,4)

, and a focal diameter of

28

.

Identify standard form: Identify the standard form of a vertical parabola. $\newline$ The standard form of a vertical parabola is $(x-h)^2 = 4p(y-k)$ , where $(h,k)$ is the vertex and $p$ is the distance from the vertex to the focus.
Determine vertex: Determine the vertex of the parabola. $\newline$ Since the parabola opens up and the focus is at $(0,4)$ , the vertex is directly below the focus. The focal diameter is $28$ , so the distance from the vertex to the focus (value of $'p'$ ) is $\frac{28}{2} = 14$ . Therefore, the vertex is at $(0,4-14) = (0,-10)$ .
Calculate value of 'p': Calculate the value of 'p'. $\newline$ The value of 'p' is half the focal diameter, so $p = \frac{28}{2} = 14$ .
Write parabola equation: Write the equation of the parabola using the vertex $(h,k)$ and the value of $p$ . Substitute $h = 0$ , $k = -10$ , and $p = 14$ into the standard form equation of a vertical parabola to get $(x-0)^2 = 4\cdot14(y+10)$ .
Simplify equation: Simplify the equation. $\newline$ The equation simplifies to $x^2 = 56(y+10)$ .

More problems from Write equations of parabolas in vertex form using the focus and directrix

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Write the equation for a parabola with a focus at \(\(-4,

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Question

10\text{-kg}

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. Initially the bucket contains

42\, \text{kg}

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14\, \text{m}

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Question

10-\mathrm{kg}

bucket is lifted from the ground to a height of

14

m at a constant speed with a rope that weighs

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. Initially the bucket contains

42

kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the

14-\mathrm{m}

level. How much work is done? (Use

9.8 \mathrm{~m} / \mathrm{s}^{2}

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Show how to approximate the required work (in J) by a Riemann sum. (Let

x

be the height in meters above the ground. Enter

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\newline

Express the work (in J) as an integral in terms of

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\newline

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Evaluate the integral (in J). (Round your answer to the nearest integer.)

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10\text{-kg}

bucket is lifted from the ground to a height of

14\text{ m}

at a constant speed with a rope that weighs

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. Initially the bucket contains

42\text{ kg}

of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the

14\text{-m}

level. How much work is done? (Use

9.8\text{ m/s}^2

g

.) Show how to approximate the required work (in

\text{J}

) by a Riemann sum. (Let

x

be the height in meters above the ground. Enter

x_i^*

x_i

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\begin{array}{l}f(t)=\left\{\begin{array}{ll}-\frac{64}{t} & , \quad t=8 \\ 14-t & , \quad t=10 \\ t^{2}-3 t+2 & , \quad t \neq 8,10\end{array}\right. \\ f(2)=\square\end{array}

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