Compute the points of the intersection between the circle x^2+y^2=1 and hyperbola xy=1?

I tried to put 1/y in place of x but I couldn't solve.

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- Apr 16th 2013, 10:49 PMkastamonuconics
Compute the points of the intersection between the circle x^2+y^2=1 and hyperbola xy=1?

I tried to put 1/y in place of x but I couldn't solve. - Apr 16th 2013, 11:02 PMGusbobRe: conics
You have $\displaystyle x^2+y^2=1$ and $\displaystyle xy=1$.

We may assume $\displaystyle x,y\not=0$ since they don't lie on the hyperbola anyways.

Taking $\displaystyle x=1/y$ as you have tried,

$\displaystyle \frac{1}{y^2}+y^2=1$

Multiplying by $\displaystyle y^2$ and rearranging gives

$\displaystyle y^4-y^2+1=0$

which has no real solutions. So there is no intersection. In fact, this is obvious if you graphed the two functions. - Apr 17th 2013, 02:58 AMkastamonuRe: conics
There is an answer. By discriminant we can find the roots. t^2=x^4 and we can find the roots.

\Delta = \,b^2-4ac. - Apr 17th 2013, 03:17 AMGusbobRe: conics
Yes, but the discriminant is negative, so all the roots are non-real numbers in the complex plane. That is to say there is no real solution. As I have suggested earlier, just graphing these two functions makes it clear that they do not intersect on the xy plane.

- Apr 17th 2013, 04:06 AMMINOANMANRe: conics
- Apr 17th 2013, 09:20 AMkastamonuRe: conics
It is positive.

- Apr 17th 2013, 09:20 AMkastamonuRe: conics
good drawing.

- Apr 17th 2013, 09:21 AMkastamonuRe: conics
- Apr 18th 2013, 08:34 PMibduttRe: conics
The equation reduces to t^2-t+1=0 when we put t = y^2.

Now discriminant of equation is b^2-4ac= (1)^2- 4*1*1= 1 - 4 = -3 NEGATIVE - Apr 19th 2013, 04:35 AMkastamonuRe: conics
ıt was x^2+y^2=4.I made a mistake.

- Apr 19th 2013, 09:02 AMMINOANMANRe: conics
Kastamonu ...........

no further comments

Attachment 28035 - Apr 19th 2013, 08:17 PMibduttRe: conics
In that case the equation reduces to t^2+4t-4=0; Now the discriminant = 4^2-4(1)(-4)= 32 so we will her discriminant = 4sqrt2

take the positive root because t = x^2 and proceed to get the required points. - Apr 19th 2013, 10:24 PMkastamonuRe: conics
many thanks.