1. ## simplify conic equation

Can someone show me how to simplify this equation, so it represents a conic section.

x^2-2x+y^2-8x+1=0

2. Originally Posted by cityismine
Can someone show me how to simplify this equation, so it represents a conic section.

x^2-2x+y^2-8x+1=0
should the -8x be - 8y?

do you know how to complete the square?

write as

x^2 - 2x + y^2 - 8y = -1

now, complete the squares for the x's and the y's separately. then you should be able to see how to rearrange it. can you continue?

3. Group and compete the square:

I assume you have a typo and mean:

$x^{2}-2x+y^{2}-8y=-1$

$(x^{2}-2x+1)+(y^{2}-8y+16)=-1+1+16$

Factor:

$(x-1)^{2}+(y-4)^{1}=16$

There....what do you have?.

4. That's exactly how it is in the book, I guess it's typo in the book.

5. Originally Posted by cityismine
That's exactly how it is in the book, I guess it's typo in the book.
yes, well, it seems likely to be a typo. you can check the answer in the back of the book, if possible, to see.

anyway, galactus gave you the answer:

$(x-1)^{2}+(y-4)^{2}=16$

what kind of conic section does this describe? maybe writing it as $(x-1)^{2}+(y-4)^{2}=4^2$ would help...

6. The answer in the book is (1,4). So I guess that's the center of the circle.

7. Originally Posted by cityismine
The answer in the book is (1,4). So I guess that's the center of the circle.
you should not be guessing. you should be able to tell from the answer galactus gave

the equation of a circle can be written in the form: $(x - h)^2 + (y - k)^2 = r^2$

when in this form, the center is $(h,k)$ and the radius is $r$