Question

What is the standard form of the equation of the conic given by 2x^2 + 2y^2 + 4x - 12y – 22 = 0?

Answer 1

@aum

Answer 2

What's next?

Answer 3

2x^2 + 2y^2 + 4x - 12y – 22 = 0
divide throughout by 2:
x^2 + y^2 + 2x - 6y - 11 = 0
(x^2 + 2x) + (y^2 - 6y) – 11 = 0
complete the square of (x^2 + 2x)
complete the square of (y^2 - 6y)

First complete the two squares and we can go to the next step.

Answer 4

How do I complete the squares?

Answer 5

Look under "steps for completing the square" here:
http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm

Answer 6

Is it (x+1)^2-1 and (y-3)^2-9

Answer 7

correct.

Answer 8

Awesome, so what's next?

Answer 9

(x^2 + 2x) + (y^2 - 6y) – 11 = 0
(x + 1)^2 - 1 + (y - 3)^2 - 9 - 11 = 0
(x+1)^2 + (y-3)^2 = 21
\((x+1)^2 + (y-3)^2 = (\sqrt{21})^2 \) Compare it to:
\((x-h)^2 + (y-k)^2 = r^2 \)
This is a circle with center at (-1, 3) and radius \(\sqrt{21}\)

Answer 10

h = -1, k = 3, r = \(\sqrt{21}\)
where (h, k) is the center of the circle and r is its radius.

Answer 11

They are just asking you to put the equation in standard form and so the answer is:
\((x+1)^2 + (y-3)^2 = (\sqrt{21})^2\)

They don't ask you to identify the equation. So the rest are all FYI.

Answer 12

Can you help with one more?