Question

Classify the conic section and write its equation in standard form.

1) y2 - 8y - 8x = -24

Answer 1

going to be a parabola
what is your version of "standard form"?

Answer 2

maybe it is \((y-k)^2=4p(x-h)\)

Answer 3

what do you mean ?

Answer 4

it says "write in standard form" so to do this we must know exactly what "standard form" means, otherwise the question makes no sense

Answer 5

i think that is the standard form

Answer 6

ok then we can do it

Answer 7

ok

Answer 8

\[y^2 - 8y - 8x = -24\] add \(8x\) get
\[y^2-8y=4x-24\] then "complete the square to get \[(y-4)^2=4x-28+16\] or
\[(y-4)^2=4x-12\]
factor on the right and get
\[(y-4)^2=4(x-3)\]

Answer 9

thanks can u help with 2 more just like this one ??

Answer 10

sure

Answer 11

x2 + y2- 4x + 2y - 11 = 0

Answer 12

for this one you have to complete the square as well
since the square terms are \(x^2\) and \(y^2\) it is a circle

Answer 13

standard form will be
\[(x-h)^2+(y-k)^2=r^2\] start by grouping terms \[x^2 + y^2- 4x + 2y - 11 = 0\]
\[x^2-4x+y^2+2y=11\]

Answer 14

then complete the square for both
\[(x-2)^2+(y+1)^2=11+4^2+1^2\]
\[(x-2)^2+(y+1)^2=29\]

Answer 15

center of the circle is \((2,-1)\) and radius is \(\sqrt{29}\)
what you need to know in order to do these is how to complete the square

Answer 16

kay thanks .

Answer 17

4x2 + y2 - 8x - 8 = 0

Answer 18

this one is an ellipse because the coefficient of the \(x^2\) and \(y^2\) terms are different

Answer 19

standard form is
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 20

\[4x^2 + y^2 - 8x - 8 = 0\]
\[4x^2-8x+y^2=8\]\[4(x^2-2x)+y^2=8\]
\[4(x-1)^2+y^2=8+4=12\]then divide by 12 and get
\[\frac{(x-1)^2}{3}+\frac{y^2}{12}=1\]

Answer 21

okay i really appreciate i have one more 4x2 - 25y2 - 100y = 0

Answer 22

this one a hyperbola because of the \(-25y^2\)

Answer 23

did you copy it correctly ?

Answer 24

4x2 - 25y2 - 100y = 0 yes

Answer 25

ok then it is nothing

Answer 26

so its not a standard form?

Answer 27

ok wait i made a mistake, hold on

Answer 28

k

Answer 29

\[4x^2 - 25y^2 - 100y = 0 \]\[4x^2-25(y^2+4)=0\]
\[4x^2-25(y+2)^2=-100\]
\[25(y+2)^2-x^2=100\]
\[\frac{(x+2)^2}{4}-\frac{x}{100}=1\]

Answer 30

typo on the last line, should be \[\frac{(y+2)^2}{4}-\frac{x}{100}=1\]

Answer 31

thank you !