# Homework Help: Inner products and Circles

1. Mar 4, 2012

### TranscendArcu

Mostly I'd like to look at the third part of the problem. I'm not sure if this is the correct way to derive the equation:

So, finding the length of a given vector given this inner product:
$<(x,y),(x,y)> = 5x^2 + y^2$.

Taking the length, we have

$|(x,y)| = \sqrt{5x^2 + y^2}$, which we define as equaling 1. Squaring both sides we find,

$5x^2 + y^2 = 1$. I think this is the equation of the circle, but I'm not sure. If it is, then my picture has y-intercepts at 1,-1 and x-intercepts at -sqrt(1/5),sqrt(1/5).

Is this correct?

2. Mar 4, 2012

### Bacle2

I think you're missing some terms from the length:

<(x,y),(x,y)>=5x2+2(xy+yx)+y2

3. Mar 4, 2012

### TranscendArcu

Whoops. You're right. My real equation is $5x^2 -2(xy+xy) +y^2 =1$. This changes shape of the circle (it's more elongated in quadrants I and III now), but the intercepts remain the same I think. No?

4. Mar 4, 2012

### Bacle2

I think if you do a rotation of the plain, you may be able to get rid of the mixed xy-terms.