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Find the vertex, focus, directrix, and focal width of the parabola. -1/40 x^2=y

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Answer:

the vertex is at (0,0)  

the focus is at (0, -10)  

the directrix is y = 10  

the focal width = |-40| = 40

Step-by-step explanation:

The equation of the parabola :

y =frac{-1}{40}x^2

Which shows that the given parabola is a vertical parabola and the standard form of the vertical parabola is written as :

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

the vertex is at (h,k)  

the focus is at (h,k+p)  

the directrix is y = k−p  

the focal width = |4p|

So, changing the given equation of parabola into standard form we get :

y =frac{1}{4times -10}(x-0)^2+0

On comparing it with equation (1) we get

p = -10 , h = 0 , k = 0

Therefore, the vertex is at (0,0)  

the focus is at (0, -10)  

the directrix is y = 10  

the focal width = |-40| = 40

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If the telescope is 1 m deep and 8 m wide, how far is the focus from the vertex? 1 2 4 16 Find the vertex, focus, directrix, and focal width of the parabola. negative 1 divided by 12 times x squared = y

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First, let me introduce the general equation of the parabola:

(x-h)^2 = +/- 4a(y-k) or (y-k)^2=+/- 4a(x-h), where

(h,k) are the coordinates of the vertex
a is the distance of the vertex to the focus
4a = length of lactus rectum or the focal width

If the equation contains (x-h)^2, then the parabola passes the x-axis twice. Similarly, (y-k)^2 passes the y-axis twice. If the sign is (-), it opens to the left(if y-axis) or downward (if x-axis). If the sign is (+), it opens to the right(if y-axis) or upward (if x-axis). 

The equation of the parabola is -1/12 x^2 = y. Rearranging to the general form:

x^2 = -12y
Therefore, 

-4a = -12
4a = 12
a = 3, and the parabola is facing downwards.

The vertex is (0,0) at the origin.
The focus is (0,-3). Since it is negative, the focus is situated downwards, hence -3.
The directrix is the mirror image of the focus. Hence, it is a line passing +3 on the y-axis. y=3
Focal width is 4a which is equal to 12 units.

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

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Standard form is, hold a sec

x=2 is directix
that means it opens left or right
so we must use
(y-k)²=4p(x-h)
where vertex is (h,k) and p is distance from focus to vertex
also shortest distance from vertex to directix
the shortest distance from focus to directix is 2p
if p>0 then the parabola opens right
if p<0 then pareabola opens left

so
(-2,0) and x=2
the distance is 4
4/2=2
p=2
wait, positive or negative
focus is to the left of the directix so p is negative
p=-2

vertex is 2 to the right of the focus and 2 to the left of directix
vertex is (0,0)

so
(y-0)²=4(-2)(x-0) or
y²=-8x is da equation
not sure what form is standard tho

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Derive the equation of the parabola with a focus at (2, –1) and a directrix of y = – .5

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Check the picture below…the parabola, with that focus point and that directrix.. .looks more or less like so

now, notice that, the directrix is above the focus, meaning the parabola is vertical and it opens downwards

the distance “p” is the same distance from the vertex to the directrix, since the vertex is half-way between those two fellows

now the  focus point is at 2,-1 and the directrix up above at +0.5, from -1 to 0.5 over the y-axis, there are 1.5 units, the vertex is half-way through, so 1.5/2 or 0.75

so the vertex is from -1 up 0.75 units, or from 0.5 down 0.75 units, that places it at -0.25, the axis of symmetry is x = 2, so, the vertex is at 2, -0.25

because the parabola is opening downwards, the value for “p” is negative, so, we know the distance “p” is 0.75, but because is an opening downwards parabola, then  p = -0.75

now, let’s plug all those guys in, let’s use for -0.25 then -1/4, and for -0.75 the -3/4, which is just the fraction version

bf textit{parabola vertex form with focus point distance}\\&#10;begin{array}{llll}&#10;(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\&#10;boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\&#10;end{array}&#10;qquad &#10;begin{array}{llll}&#10;vertex ({{ h}},{{ k}})\\&#10;{{ p}}=textit{distance from vertex to }\&#10;qquad textit{ focus or directrix}&#10;end{array}\\

bf -------------------------------\\&#10;(x-{{ h}})^2=4{{ p}}(y-{{ k}}) qquad &#10;begin{cases}&#10;h=2\&#10;k=-frac{1}{4}\&#10;p=-frac{3}{4}&#10;end{cases}&#10;\\\&#10;(x-2)^2=4left( -frac{3}{4} right)left( y-left( -frac{1}{4} right) right)implies (x-2)^2=-3left( y+frac{1}{4} right)&#10;\\\&#10;cfrac{(x-2)^2}{-3}=y+cfrac{1}{4}implies cfrac{(x-2)^2}{-3}-cfrac{1}{4}=y&#10;\\\&#10;-cfrac{1}{3}(x-2)^2-cfrac{1}{4}=y

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