1. ## conics

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.

2. ## Re: conics

Originally Posted by kastamonu
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.
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.

3. ## Re: 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.

4. ## Re: 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.

6. ## Re: conics

It is positive.

7. ## Re: conics

good drawing.

8. ## Re: conics

Originally Posted by Gusbob
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.
It is positive.

9. ## Re: 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

10. ## Re: conics

ıt was x^2+y^2=4.I made a mistake.

11. ## Re: conics

Kastamonu ...........