Question

x^2-y^2=x+y+1. I know this is a hyperbola, but can u put it in the standard equation

Answer 1

complete the squares.

Answer 2

give me a second

Answer 3

i got
\[\large (x-1/2)^2-(y+1/2)^2=1 \]

Answer 4

how did u get the one

Answer 5

i will do all the steps
\[\large x^2-y^2=x+y+1 \]
\[\large x^2-x-y^2-y=1 \]
\[\large (x^2-x+(1/2)^2)-(y^2+y+(1/2)^2)=1+(1/2)^2-(1/2)^2=1 \]
\[\large (x-1/2)^2-(y+1/2)^2=1 \]

Answer 6

thank u, I forgot to subtract the 1/4 so I ended up with 3/2 on the left. that is where I got stuck

Answer 7

u r welcome

Answer 8

I was right that it is a hyperbola

Answer 9

yes

Answer 10

my friend and I out this ellipise in its standard form x^2+4y^2=2x+8y-1. she got this answer (x-1)^2 + 4(y-1)^2 =4. I got (x-1)^2/(16) + (y-4)^2/(4)= 1. which is correct

@helder_edwin

Answer 11

give a minute

Answer 12

i got the first one:
\[\large \frac{(x-1)^2}{4}+(y-1)^2=1 \]

Answer 13

how

Answer 14

\[\large x^2+4y^2=2x+8y-1 \]
\[\large (x^2-2x+1)+4(y^2-2y+1)=-1+1+4=4 \]
\[\large (x-1)^2+4(y-1)^2=4 \]
\[\large \frac{(x-1)^2}{4}+(y-1)^2=1 \]

Answer 15

compute limit (as t approaches 0) : sec^2t -1/ (t). I got 0, just want to confirm

Answer 16

i got zero.

Answer 17

compute limit (as t approaches 0) : sect -1/ (t). I got 0 too, just want to confirm

Answer 18

I used the derivative. did u

Answer 19

if not can u show me the other way u used

Answer 20

i got zero.
yes i used L'Hôpital's rule

Answer 21

my teacher want us to find the derivative of this problem using the limit rule, but I get lose using the limit rule: (1/(x+1))