Equation of Conics

Question

Write the equation of the conic in standard form and indicate which conic it is 

x^2 + 4y^2 - 6x + 16y + 21=0

 

This is what I did 

 

(x^2 - 6x + 9) - 4(y^2 + 4y +4)=-21 + 9 - 4

 

(x-3)^2  -  4(y+2)^2 =-16

 

((x-3)^2/-16) - (4(y+2)^2/-16) = -16/-16

 

Don't know if this is the answer :

((x-3)^2/-16) - ((y+2)^2/-4) = 1

Answer 1

Is there a negative in the original problem?  In front of the y squared term?  Because I don't know where that came from in your solution.  Also, you would need to factor out the 4 from ALL of your y terms, so when you complete that square, you would still have your (y+2), but you would have added 4*4, or 16 on the other side.  You are off to a nice start.

 

Here's my work and solution for the original problem as written:

 

x²-6x+9+4(y²+4y+4) = -21+9+16

(x-3)²+4(y+2)²=4

 

(x-3)²/4 + (y+2)²/1 = 1