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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- April 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. - April 16th 2013, 11:02 PMGusbobRe: conics
- April 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. - April 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.

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

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

- April 17th 2013, 09:21 AMkastamonuRe: conics
- April 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 - April 19th 2013, 04:35 AMkastamonuRe: conics
ıt was x^2+y^2=4.I made a mistake.

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

no further comments

Attachment 28035 - April 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. - April 19th 2013, 10:24 PMkastamonuRe: conics
many thanks.