I need help in answering a short quiz on pre calc

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Itโ€™s only 2 pages with 7 questions 4 multiple choices and 3 graphing questions

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Graph the equation List the center, vertices and foci. 6) (x + 2)2. y - 12 16 9 || - 10 10 x Center Vertices Foci Graph the equation. List the center, vertices, foci and asymptotes. (x + 2)2 (y - 12 7) -1 4 25 101 || -10 10 Center Vertices Foci Asymptotes 2 Pre Calculus Algebra Quiz 4 Name Name the conic. If the conic is a circle then indicate circle and not ellipse for the answer 1) 4x2 + 8x - 5y2 - 15y = 2 A) parabola B) ellipse C) hyperbola D) circle 2) 4x2 + 12x + 6y2.127 - 24 A) parabola B) circle C) ellipse D) hyperbola 3) 2 - 7y - 3x+12 A) circle B) ellipse C) parabola D) hyperbola D) ellipse 4) 5x2 + 9x + 5y2 - 97 - 15 A) circle B) hyperbola C) parabola Find the vertex, focus, and directrix and end points of the latus rectum. Graph the equation. 5) () - 2)2 . -8(x + 1) 105 -10 +++ 10 x 5 Vertex Focus Directrix Endpoints of the Latus Rectum 1
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Explanation & Answer

Done.Here's the solution file (both in doc and pdf format).๐Ÿ˜Š

Name the conic. If the conic is a circle then indicate circle and not ellipse for
the answer.
(1) 4๐‘ฅ 2 + 8๐‘ฅ โˆ’ 5๐‘ฆ 2 โˆ’ 15๐‘ฆ = 2
A) Parabola
B) Ellipse
(2) 4๐‘ฅ 2 + 12๐‘ฅ + 6๐‘ฆ 2 โˆ’ 12๐‘ฆ = 24

C) Hyperbola

D) Circle

A) Parabola
B) Circle
2
(3) ๐‘ฆ โˆ’ 7๐‘ฆ = 3๐‘ฅ + 12

C) Ellipse

D) Hyperbola

A) Circle
B) Ellipse
2
2
(4) 5๐‘ฅ + 9๐‘ฅ + 5๐‘ฆ โˆ’ 9๐‘ฆ = 15

C) Parabola

D) Hyperbola

A) Circle
B) Hyperbola
C) Parabola
D) Ellipse
(5) Find the vertex, focus, directrix and endpoints of the latus rectum. Graph
the equation.
(๐‘ฆ โˆ’ 2)2 = โˆ’8(๐‘ฅ + 1)
Sol: Vertex:
The parabola having vertex (h, k) in standard form is
(๐‘ฆ โˆ’ ๐‘˜)2 = 4๐‘(๐‘ฅ โˆ’ โ„Ž)
To find the vertex (h, k), we need to write the equation in standard form of
parabola. So,
(๐‘ฆ โˆ’ 2)2 = 4(โˆ’2)(๐‘ฅ โˆ’ (โˆ’1))
Comparing the equation with standard form equation, we have
โ„Ž = โˆ’1, ๐‘˜ = 2, ๐‘ = โˆ’2
Hence
Vertex = (โˆ’1, 2)
Directrix:
The parabola of the form
(๐‘ฆ โˆ’ ๐‘˜)2 = 4๐‘(๐‘ฅ โˆ’ โ„Ž)
has the directrix ๐‘ฅ = โ„Ž โˆ’ ๐‘.

To find the directrix, we again need to write the equation of parabola in
standard form which is
(๐‘ฆ โˆ’ 2)2 = 4(โˆ’2)(๐‘ฅ โˆ’ (โˆ’1))
Comparing the equations, we get
โ„Ž = โˆ’1, ๐‘˜ = 2, ๐‘ = โˆ’2
So, to find the directrix, we use the equation ๐‘ฅ = โ„Ž โˆ’ ๐‘ for this form of
parabola,
๐‘ฅ = โ„Žโˆ’ ๐‘
Putting the values, we get
๐‘ฅ = โˆ’1 โˆ’ (โˆ’2)
๐‘ฅ = โˆ’1 + 2
๐‘ฅ=1
Hence the directrix is ๐‘ฅ = 1
Endpoints of the latus rectum:
First, we need to find the focus.
The focus of parabola of the form
(๐‘ฆ โˆ’ ๐‘˜)2 = 4๐‘(๐‘ฅ โˆ’ โ„Ž)
is (h+p, k)
Considering the equation again in standard form, we get
(๐‘ฆ โˆ’ 2)2 = 4(โˆ’2)(๐‘ฅ โˆ’ (โˆ’1))
Here, โ„Ž = โˆ’1, ๐‘˜ = 2, ๐‘ = โˆ’2
So, putting the values, we have the focus
(โˆ’1 + (โˆ’2),2)
(โˆ’1 โˆ’ 2,2)
(โˆ’3,2)
To find the end points of latus rectum, we need to know the length of latus
rectum. The length of latus rectum is |4p|.

So, the length of latus rectum is
๐‘™ = |4๐‘| = |4(โˆ’2)| = | โˆ’ 8| = 8
Now, the endpoints of latus rectum for this type of parabola can be found
using the focus point as
๐‘™

8

2

2

First endpoint can be found by moving = = 4 units up in the direction of
positive y axis from y coordinate of the focus and second point can be found
moving 4 unit down in the direction of negative y axis. So,
๐‘ƒ๐‘œ๐‘–๐‘›๏ฟฝ...


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