# Conics Problem Explanation

Problem Statement:

Circle $\Gamma$ intersects the hyperbola $y = \frac 1x$ at $(1,1), \left(3,\frac13\right)$, and two other points. What is the product of the $y$ coordinates of the other two points?

My Work:

$x^2+y^2+2gx+2fy+c=0$ and substitute x=$\dfrac{1}{y}$ to get 4th degree equation in y. $y^4+..... + 1=0$ so product of all roots =1. Known roots: 1 and $\dfrac {1}{3}$. So the product of the two other roots will be $\dfrac{1}{\dfrac{1}{3}}=3$.

How would I justify the steps in my proof? How would I make my proof more "complete"?

• No need to say more. One thing you can say just before obtaining your 4th degree equation is "ordinates of intersection points are the roots of the equation obtained by plugging x=1/y in the circle's equation" – Jean Marie Jan 31 '17 at 22:45

The circle $\Gamma$ must have an equation of the form $x^2 + y^2 + 2gx + 2fy + c = 0$ for some $c, f, g$. The points of intersection must also satisfy $x = 1/y$, and therefore $$\left(\frac 1y\right)^2 + y^2 + 2g\left(\frac 1y\right) + 2fy + c = 0$$ So $$y^4 + 2fy^3 + cy^2 + 2gy + 1 = 0$$ This equation factors to $$(y - y_1)(y - y_2)(y - y_3)(y - y_4) = 0$$ for the $y$-values of the four points of intersection, $y_1, y_2, y_3, y_4$. If we multiply that back out, we see that the constant term is $y_1y_2y_3y_4 = 1$.
Two of those values are given as $y_1 = 1$ and $y_2 = \frac 13$. Thus $(1)\left(\frac 13\right)y_3y_4 = 1$, and so $$y_3y_4 = 3$$