Mathematics, 29.10.2019 07:31 gabbycasey102

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.

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Answer from: zodiacpumpkin1126

The equation of parabola is (x+5)^2=12(y-2) .

Explanation:

It is given that the focus of the parabola is (−5, 5) and a directrix of y = −1.

The standard form of a parabola is,

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

Where, (h,k+p) is the focus and y=k-p is the directrix.

It is given that the focus of the parabola is (−5, 5).

(h,k+p)=(-5,5)

On comparing both sides we get,

h=-5

k+p=5 .... (1)

The directrix of y = −1.

k-p=-1 .... (2)

Add equation (1) and (2),

2k=4

k=2

Put this in equation (1),

p=3

Put p=3, k=2 and h=-5 in standard equation of parabola.

(x-(-5))^2=4(3)(y-2)

(x+5)^2=12(y-2)

Therefore, the equation of parabola is (x+5)^2=12(y-2) .

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.

Answer from: jaueuxsn

Check the picture below.

so, the left-part of the picture is the focus point and the directrix... .so.. notice, the focus is above the directrix, meaning is a vertical parabola and is opening upwards.

now, looking at the right-part of the picture, bear in mind that, the vertex is "p" units away from the directrix and the focus, so, the vertex is half-way between those fellows, notice in the picture what "p" is, keep in mind that, because the parabola is opening upwards, "p" is a positive unit, thus is 3.

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=-5\\
k=2\\
p=3
\end{cases}\implies [x-(-5)]^2=4(3)(y-2)
\\\\\\
(x+5)^2=12(y-2)\implies \cfrac{(x+5)^2}{12}=y-2\implies \cfrac{(x+5)^2}{12}+2=y
\\\\\\
\cfrac{1}{12}(x+5)^2+2=y$

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.

Answer from: Quest

what

Answer from: Quest

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

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