Question

This is the conics
when I am given with the equation like 16y^2-x^2+2x+64y+63=0, I know how to do everything else, but when I complete the square and make it into standard form, the right side becomes 0, not 1. In this case, what should I do?

Answer 1

I'll take a stab at it

Answer 2

\[16y ^{2}-x ^{2}+2x+64y+63=0\]

Answer 3

\[16\left( y ^{2} +4y+4\right)-\left( x ^{2}+2x+1 \right)=-63+64+1\]

Answer 4

yes I got that :(

Answer 5

\[\frac{ 16 }{ 2 }(y+2)^{2}-\frac{ (x+1)^{2} }{ 2 } = 1\]

Answer 6

ohoh hold on

Answer 7

how did you get +1 ?

Answer 8

\[\frac{ (y+2)^{2} }{ \frac{ 2 }{ 16 }}-\frac{ (x+1)^{2} }{ 2 } = 1\]

Answer 9

sorry for −63+64+1, how did you get 1? isn't it -1?

Answer 10

-63+64 = 1

Answer 11

then 1+1 = 2

Answer 12

then divide both sides by 2

Answer 13

okay Thank you! :)

Answer 14

You're welcome