Find the vertex, focus, directrix, and focal width of the parabola.
1. Consider the parabolic equation: The parabolic equation is given by: …..[A] where, |4p| represents the focal width of the parabola Focus = (h, k+p) Vertex = (h, k) Directive (y) = k -p Comparing the given equation with the equation [A] We have; We have; 4p = 20 Divide both sides by 4 terms; p = 5 Vertex = (0,0) Focus = (0, 0+5) = (0, 5) Focal width = 20 Guideline: y = kp = 0-5 = -5 ⇒y = -5 Therefore, only option A is correct 2. Given the parabolic equation: divide both sides by 3 terms; The equation of the parabola is given by: ….[B] Vertex = (h, k) Focus = (h+p, k) directrix: x = k -p Focal width = 4p Comparison of the given equation with the equation [B] We have; Divide both sides by 4 terms; Focal width = Vertex = (0, 0) Focus = guideline: ⇒ Therefore option B is correct. 3. The equation of the parabola that opens is: For the given problem: Axis of symmetry: x = 0 Distance from a focus to the vertex on the axis of symmetry: p = 9 then; 4p = 36 ⇒ Divide both sides by 36 terms; Therefore, only option A is correct. 4. The equation of the parabola that opens is: Given that: Focus = (0, 8) and directrix: y = -8 The distance from the focus to the vertex and from the vertex to the directrix is the same: i, e |p | = 8 So 4p = 32 ⇒ Dividing both sides by 32 we have; Therefore, only option A is correct. 5. The equation of the parabola that opens to the right is: Given that: Focus: (7, 0) and directrix: x = -7 The distance from the focus to the vertex and from the vertex to the directrix is the same : i,e |p| = 7 then 4p = 28 ⇒ Dividing both sides by 32 we have; Therefore, option B is correct. 6. According to the statement: A building has a parabolic arch entrance 74 feet high and 28 feet wide at the base, as shown below. The equation of the parabola is given by: …..[C] Substitute the point (14, -74) we have; Put x=14 and y=-74 then; ⇒ Divide both sides by 74 terms; Substitute in the equation [C] We have; or Therefore, an equation for the parabola if the vertex is placed at the origin of the coordinate system is,
A Walkthrough: I Just Passed the Edge Test
option:A Vertex: (0,0) Focus: (0,-4) Directing line: y=4 focal width: 16 Step by step explanation: Part 1: we are given an equation of a parabola as: when conversion of the parabola equation in the standard form of: we have the vertex as (h,k). so here the equation is converted as follows: we have h=0,k=0 and p= -4 so we have the vertex as (0,0). now the focus is given by (h,k+p) so the focus is :(0,-4) the directrix is given by the formula y=kp Therefore the directrix is_y=4. the focal width of a parabola is given by |4 p| So the focal width is: 16. So option A is correct.
option:A Vertex: (0,0) Focus: (0,-4) Directing line: y=4 focal width: 16 Step by step explanation: Part 1: we are given an equation of a parabola as: when conversion of the parabola equation in the standard form of: we have the vertex as (h,k). so here the equation is converted as follows: we have h=0,k=0 and p= -4 so we have the vertex as (0,0). now the focus is given by (h,k+p) so the focus is :(0,-4) the directrix is given by the formula y=kp Therefore the directrix is_y=4. the focal width of a parabola is given by |4 p| So the focal width is: 16. So option A is correct.
Answer 6
Part 1) Option C) y = minus one divided by thirty-six x² Part 2) Option A) y = one divided by thirty-six x² Part 3) Vertex: (0, 0); Focus: one divided by sixteen point zero; Guideline: x = minus one divided by sixteen; Focal width: 0.25 Part 4) Option C) x = one divided by twelve y² Walkthrough: Part 1) Find the standard form of the parabola equation with a focus on (0, -9) and a directrix y = 9 We know that the shape of the vertex of the equation of the vertical parabola is equal to where Vertex —-> (h,k) Focus —-> F(h,k+p) directrix —–> y= kp we have F(0, -9) then h =0 k+p=-9 —–> equation A y=9 then kp=9 —-> equation B Add equation A and equation B k+p =-9 kp=9 ——— – 2k=0 k=0 then Find the value of p 0+p=-9 p=-9 replace in the equation Convert to standard form isolate the variable y Part 2) Find the standard form of the parabola equation with vertex at the origin and focus at (0, 9) we know that the form of the vertex of the vertical parabola equation is equal to where Vertex —-> (h ,k) Focus —-> F(h, k+p) directrix —–> y=kp we have the vertex (0,0) —–> h=0,k=0 F(0,9) then k+ p=9——> 0+p=9 — –> p=9 replace in equation Convert to standard form is ol the variable y Part 3) Find e and the vertex, focus, directrix and width f locus of the parabola. x = 4y² we know that The shape of the vertex of the equation of the horizontal parabola is equal to where Vertex —-> (h,k) Focus —-> F(h+p,k) directrix —–> x=hp we have —–> then Vertex (0,0) ——> h=0,k=0 4p=1/4 ——> Focal width p=1/16 Focus F(0+1/16,0) —- > F (1/ 16.0) directive —–> x=0-1/16 —–> x=-1/16 so Vertex: (0, 0); Focus: one divided by sixteen point zero; Guideline: x = minus one divided by sixteen; Focal width: 0.25 Part 4) Find the standard form of the parabola equation with focus at (3, 0) and directrix at x = -3. we know that The shape of the vertex of the equation of the horizontal parabola is equal to where Vertex —-> (h,k) Focus —-> F(h+p,k) directrix —–> x=hp we have Focus F( 3.0 ) then h+p=3 ——> equation A k=0 principal x=-3 then hp=-3 ——> equation B Add equation A and equation B and solve for h h+p =3 hp= – 3 ———— 2h=0 —–> h=0 Find the value of p h+p=3 ——> 0+p=3 ——> p=3 replace in equation Convert to standard form isolate the variable X
Answer 7
the answer is the letter C Step by step explanation:
Vertex: (0, 0); Focus: (0, -4); Guideline: y = 4; Focal length: 16 ⇒ answer (a) Walkthrough: * Let’s review some parabola facts – Standard form equation for a parabola with vertices at (0 , 0) – If the equation is in the form x² = 4py , then – The axis of symmetry is the y axis, x = 0 – 4p is equal to the coefficient of y in the equation given to solve p – If p > 0, the parabola opens. – If p < 0, the parabola opens downwards. - Use p to find the focus coordinates, (0, p) - Use p to find the governing equation, y= − p - Use p to find the endpoints of the focal diameter, (±2p, p) * Now let's solve the problem - The vertex of the parabola is (0 , 0) ∵ -1/16x² = y ⇒ multiply each side by -16 ∴ x² = -16y ∴ 4p = -16 ⇒ ÷ 4 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 ends of the diameter focal, (±2p , p) ∴ Focal width = 2p - (-2p) = 4p ∴ Focal width = 4 × I-4I = 16 * Vertex: (0, 0); Focus: (0, -4); Guideline: y = 4; Focal width: 16
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