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Anonymous

Feb 20, 2025

Find the vertex, focus, directrix, and focal width of the parabola.

Could you please help me find the vertex, focus, directrix, and focal width of the given parabola? Your detailed explanation will be greatly appreciated.

10 Answers

A
Anonymous

Dec 11, 2024

1) We have that the equation is , hence y=x^2/20. The standard equation of such an equation is y=. Hence, p=5 in this case. The focus is at (0,5) and the directrix is at y=-5 (a tip is that the directrix is always "opposite" the focus point of a parabola; if the directrix is at x=-7 for example, the focus is at (7,0)).2) Similarly, we have that the equation is . Thus, p=1/12. In this case, the parabola opens along the x-axis and the focus is at (1/12, 0). Also, the directrix is at x=-1/12. Hence the correct answer is B.3) We are given that the parabola has a p of 9. Also, the focus lies along the y-axis, hence the parabola is opening along the y-axis. Finally, the focus is on the positive half, so the parabola is opening upwards. The equation for this case is y=.4) Similarly as above. The directrix is superfluous, we only need the p-value. THe same comments about the parabola apply and if we substitute p=8 in the formula: we get y=.5) This is somewhat different, even though we do not need the directrix again. The focus lies on the x-axis, thus the parabola opens in this direction. The focus lies on the positive part of the axis, thus the parabola opens to the right. We also are given p=7. Hence, the equation we need is of the form. Substituting p=7, we get .6) The equation of a prabola with a vertex at (0,0) is of the form y=-ax^2. The minus sign is needed since the parabola is downwards. Since we are given anothe point, we can calculate a. We have to take y=-74 and x=14 feet (since left to right is 28, we need to take half). . Thus a=0.378. Hence the correct expressions is y=-0.378*
A
Anonymous

Jan 30, 2025

1.Given the parabolic equation: The equation of parabola is given by:                       .....[A]where,|4p| represents the focal width of the parabolaFocus = (h, k+p)Vertex = (h, k)Directrix (y) = k -pOn comparing given equation with equation [A] we have;we have;4p = 20 Divide both sides by 4 we have;p = 5Vertex =(0,0)Focus = (0, 0+5) = (0, 5)Focal width = 20Directrix:y = k-p = 0-5 = -5⇒y = -5Therefore, only option A is correct2. Given the parabolic equation:Divide both sides by 3 we have;The equation of parabola is given by:             ....[B]Vertex = (h, k)Focus = (h+p, k)directrix: x = k -pFocal width = 4p Comparing given equation with equation [B] we have; Divide both sides by 4 we have;Focal width = Vertex = (0, 0)Focus = directrix:⇒Therefore, option B is correct.3.The equation of parabola that opens upward is:For the given problem:Axis of symmetry:x = 0Distance from a focus to the vertex on the axis of the symmetry:p = 9then;4p = 36⇒Divide both sides by 36 we have;Therefore, the only option A is correct.4.The equation of parabola that opens upward is:Given that: Focus = (0, 8) and directrix: y = -8Distance from a focus to the vertex and vertex to directrix is same:i,e|p| = 8Then, 4p = 32⇒Divide both sides by 32 we have;Therefore, the only option A is correct.5.The equation of parabola that opens right is:Given that:Focus: (7, 0) and directrix: x = -7Distance from a focus to the vertex and vertex to directrix is same:i,e|p| = 7then4p = 28⇒Divide both sides by 32 we have;Therefore, the option B is correct.6.As per the statement:A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base as shown below.The equation of parabola is given by:            .....[C]Substitute the point (14, -74) we have;Put x = 14 and y = -74then;⇒Divide both sides by 74 we have;Substitute in the equation [C] we have;orTherefore,  an equation for the parabola if the vertex is put at the origin of the coordinate system is,
A
Anonymous

Jan 11, 2025

AStep-by-step explanation:i just took the edg quiz
C
Chelsey Roob

Jan 20, 2025

option:AVertex: (0,0)Focus: (0,-4)Directrix: y=4focal width: 16Step-by-step explanation:Part 1:we are given a equation of a parabola as:on converting the equation of parabola into the standard form of: we have vertex as (h,k).so here the equation is converted as:we have h=0,k=0 and p= -4so, we have vertex as (0,0).now the focus is given as (h,k+p)hence focus is:(0,-4)directrix is given by the formula y=k-pHence, directrix is:y=4.focal width of a parabola is given by |4 p|Hence, Focal width is: 16.Hence, option A is correct.
M
Marco Littel

Feb 19, 2025

option:AVertex: (0,0)Focus: (0,-4)Directrix: y=4focal width: 16Step-by-step explanation:Part 1:we are given a equation of a parabola as:on converting the equation of parabola into the standard form of: we have vertex as (h,k).so here the equation is converted as:we have h=0,k=0 and p= -4so, we have vertex as (0,0).now the focus is given as (h,k+p)hence focus is:(0,-4)directrix is given by the formula y=k-pHence, directrix is:y=4.focal width of a parabola is given by |4 p|Hence, Focal width is: 16.Hence, option A is correct.
A
Anonymous

Jan 30, 2025

Part 1) Option C)  y = negative one divided by thirty six x²Part 2) Option A) y = one divided by thirty six x²Part 3) Vertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = negative one divided by sixteen; Focal width: 0.25Part 4) Option  C) x = one divided by twelve y²Step-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 thatThe vertex form of the equation of the vertical parabola is equal towhereVertex ----> (h,k)Focus ----> F(h,k+p)directrix ----->  y=k-pwe haveF(0,-9)soh=0k+p=-9 -----> equation Ay=9sok-p=9 ----> equation BAdds equation A and equation Bk+p=-9k-p=9-----------2k=0k=0soFind the value of p0+p=-9p=-9substitute in the equationConvert to standard formisolate the variable yPart 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 thatThe vertex form of the equation of the vertical parabola is equal towhereVertex ----> (h,k)Focus ----> F(h,k+p)directrix ----->  y=k-pwe haveVertex (0,0) -----> h=0,k=0F(0,9)sok+p=9------> 0+p=9 -----> p=9substitute in the equationConvert to standard formisolate the variable yPart 3)  Find the vertex, focus, directrix, and focal width of the parabola.  x = 4y²we know thatThe vertex form of the equation of the horizontal parabola is equal towhereVertex ----> (h,k)Focus ----> F(h+p,k)directrix ----->  x=h-pwe have -----> soVertex (0,0) ------> h=0,k=04p=1/4 ------> Focal widthp=1/16Focus F(0+1/16,0) ----> F(1/16,0)directrix -----> x=0-1/16 -----> x=-1/16thereforeVertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = negative one divided by sixteen; Focal width: 0.25Part 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 thatThe vertex form of the equation of the horizontal parabola is equal towhereVertex ----> (h,k)Focus ----> F(h+p,k)directrix ----->  x=h-pwe haveFocus F(3,0)soh+p=3 ------> equation Ak=0directrix x=-3soh-p=-3 ------> equation BAdds equation A and equation B and solve for hh+p=3h-p=-3------------2h=0 -----> h=0Find the value of ph+p=3 ------> 0+p=3 ------> p=3substitute in the equationConvert to standard formisolate the variable x
A
Anonymous

Dec 10, 2024

the answer is letter CStep-by-step explanation:
A
Anonymous

Jan 25, 2025

Vertex: (0, 0); Focus: (0, -4); Directrix: y = 4; Focal width: 16 ⇒ answer (a)Step-by-step explanation:* Lets revise some facts about the parabola- Standard form equation for a parabola of vertex at (0 , 0)- If the equation is in the form  x² =  4py, then  - The axis of symmetry is the y-axis,  x = 0  - 4p  equal to the coefficient of y in the given equation to   solve for  p - If  p  >  0, the parabola opens up. - If  p <  0, the parabola opens down.- Use  p  to find the coordinates of the focus,  (0  ,  p)- Use  p  to find equation of the directri ,   y= − p- Use  p  to find the endpoints of the focal diameter,  (±2p  ,  p)* Now lets solve the problem- The vertex of the parabola is (0 , 0)∵ -1/16x² = y ⇒ multiply each side by -16∴ x² = -16y∴ 4p = -16 ⇒ ÷ 4 tbe both sides∴ p = -4∵ The focus is (0 , p)∴ The focus is (0 , -4)∵ The directrix is y = -p∴ The directrix is y = -(-4) = 4 ⇒ y = 4∵ The endpoints of the focal diameter,  (±2p  ,  p)∴ The focal width = 2p - (-2p) = 4p∴ The focal width = 4 ×  I-4I = 16* Vertex: (0, 0); Focus: (0, -4); Directrix: y = 4; Focal width: 16
A
Anonymous

Dec 20, 2024

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A
Anonymous

Jan 14, 2025

Dude if you figured this out pls help me pls

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