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

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*... Show More

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,... Show More

AStep-by-step explanation:i just took the edg quiz... Show More

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.... Show More

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.... Show More

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... Show More

the answer is letter CStep-by-step explanation:... Show More

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... Show More

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Dude if you figured this out pls help me pls... Show More