Question

Help

Answer 1

B

Answer 2

@Vocaloid

Answer 3

because left would be the x-axis and going up would be the y axis

Answer 4

for parabolas written in this form, it's the opposite of what we normally expect

12y = x^2
y = (1/12)x^2

12(y+1) = x^2 becomes 12y + 12 = x^2 which becomes 12y = x^2 - 12 which becomes

y = (12)(x^2 - 1) a downward shift, the opposite of what we expected from adding 1 to y

so in order to shift 2 up, we need to subtract 2 from y
and to shift right we need to subtract 1 from x

so C not B

Answer 5

I would just work with the 5 and -x right?

Answer 6

whenever you take the square root of something you get +A, - A, or 0, making the sum 0

Answer 7

first you have to double the second equation and eliminate the x^2 terms, then half the second equation to eliminate the y^2 terms, then you'll be left with y to solve for

Answer 8

(-4,-3)

Answer 9

yeah nvm that's right

Answer 10

Thanks

Answer 11

0?

Answer 12

good

Answer 13

can we actually go back to this problem for a sec, I plugged in your answer and it didn't end up working :S

Answer 14

2x^2 + y^2 = 33
2x^2 + 2y^2 + 4y = 38

y^2 + 4y = 5
after we factor this we get y = 1 and y = -5, then we need to plug this back into the equation to get the possible x values

Answer 15

plugging in y = -5 gives x = -2, making the quadrant III solution (-2,-5)

Answer 16

Okay thanks

Answer 17

4?

Answer 18

good

Answer 19

B

Answer 20

it's a parabola and a circle, so w/o doing any calculations the max is theoretically 4 not 3

Answer 21

parabola + an ellipse technically

Answer 22

and 3 for this one right

Answer 23

yes

Answer 24

good

Answer 25

B

Answer 26

good

Answer 27

solve the first equation for y
y = 25 - x
then plug this in for y into the first equation to solve for x

Answer 28

(13,12)

Answer 29

good

Answer 30

(4,3)

Answer 31

that's quadrant I

Answer 32

sorry (4,-3)

Answer 33

good

Answer 34

3

Answer 35

check again, how many times do the red lines/blue lines cross each other?

Answer 36

4

Answer 37

good, so 4 solns not 3

Answer 38

0?FR?

Answer 39

0?

Answer 40

the first time I did this I used software and got 2,2, -2,-2 as the x-coordinates making the sum 0

we could also do it by hand if you want to confirm

Answer 41

Yeah i did it an got 0 as well

Answer 42

Lol wrong picture

Answer 43

subtracting the two eqn's gives us

4x^2 = 16

x = +/- 2 making 0 the sum :S

Answer 44

yes so it is 0

Answer 45

yeah, 0

Answer 46

(-2,3)?

Answer 47

quadrant I so both coordinates have to be positive

Answer 48

(2,3))

Answer 49

good

Answer 50

1. ellipse

Answer 51

good

Answer 52

2. hyperbola
3.

Answer 53

you'd have to complete the square for #3 to figure out what shape it is

Answer 54

3.circle

Answer 55

oh it wouldn't be circle?

Answer 56

yeah it's a circle

Answer 57

4. parabola
5. hyperbola

Answer 58

awesome, that's it

Answer 59

\[center: (4,2) \]

Answer 60

vertices (10,2)(-2,2)

Answer 61

\[Foci: (4+\sqrt[3]{5},2)\]

Answer 62

\[Foci: (4\sqrt[-3]{5},2)\]

Answer 63

\[Eccentricity: \frac{ \sqrt{5} }{ 2 }\]

Answer 64

\[assymptotes: y=\frac{ x }{ 2 }, y=-\frac{ x }{ 2 }+4\]

Answer 65

yeah so far so good (normally I would prefer to do the calculations by hand but it's pretty tedious tbh)

Answer 66

Yeah no its fine i did 3 out 6 by hand

Answer 67

I'm sure they are all correct

Answer 68

yeah I just checked and it looks fine

Answer 69

length of transverse axis

Answer 70

how do i find the length for the transverse axis and conjugate axis

Answer 71

the bigger denominator (36) is the a^2 value
the smaller denominator (9) is the b^2 value

transverse axis = 2a
conjugate = 2b

so transverse = 12 and conjugate = 6

Answer 72

perfect thanks

Answer 73

okay so i graphed it already, and got the center, vertices, and foci

Answer 74

Need help with the eccentricity , major axis , and minor axis

Answer 75

major axis is double the square root of the bigger number

so 2sqrt(16) = 2*4 = 8

minor axis is double the square root of the smaller denominator

so 2sqrt(4) = 2*2 = 4

the minor axis is 4, which is under x so the minor axis is along the x-axis

major axis is along the y-axis

Answer 76

okay so major : 8 minor: 4

Answer 77

eccentricity = c/a

a is the half length of the major axis (so a = 4)
b is half the length of the minor axis (so b = 2)

c^2 = a^2 - b^2

find eccentricity

Answer 78

2^2

Answer 79

4^2 - 2^2 = ?

remember exponents come first in pemdas

Answer 80

16-4

Answer 81

12

Answer 82

good, so c^2 = 12 and c = sqrt(12) or 2sqrt(3)

so c/a = eccentricity = 2sqrt(3) / 4, or sqrt(3)/2

Answer 83

\[\frac{ \sqrt{3} }{ 2 }\]

Answer 84

yeah, that

Answer 85

for this i got everything except for focal chord

Answer 86

brb going to teach myself what the hell a focal chord is

Answer 87

Lmao sure thing

Answer 88

alright what did you get for your p-value? b/c the focal chord is just 4p

Answer 89

honestly i just figured out vertex, etc by online calculator

Answer 90

that's fine, we can just do a quick complete the square

y^2 + 6y- 2x + 13 = 0

adding 9 to each side:

y^2 + 6y + 9 - 2x + 13 = 9

(y+3)^2 = 2x - 4 = 2(x-2)

then

if y^2 = 4px

then 4px = 2(x-2)

making 4p = 2 and p = 1/2 making the focal chord 4p = 2 = your answer for focal chord

Answer 91

Thank you so much

Answer 92

I already got #10 , for some reason I'm not getting a correct answer for #11

Answer 93

gonna get back to this asap

Answer 94

Sure take your time

Answer 95

alright, you can subtract equation 2 - equation 1 to get

2x + 4y + 12 = 0

dividing by 2:

x + 2y + 6 = 0

so x = -2y - 6

then we can just plug this into the first equation to get y

Answer 96

x^2 + y^2 + 2x + 2y = 0

( -2y - 6)^2 + y^2 + 2( -2y - 6) + 2y = 0

shoving this into a calculator gives us

y = -12/5, y = -2

then we just plug back into equation 1 (one at a time) to find the matching x-values

Answer 97

it's a lot of algebra but I end up getting

(-2,-2) and (-6/5, - 12/5) and they're both circles (since the question asks)

Answer 98

Thank you so muchh