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"?

  • 1
    $\begingroup$ 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" $\endgroup$ – Jean Marie Jan 31 '17 at 22:45

As JeanMarie indicates, you've included everything needed. The only improvements are to say it a little more clearly. As you indicate in the question, what you have written down is your work. I.e., the steps you took in your head, where you know what is going on, to solve the problem. Now you want to explain in such a way as to make the steps and why you are taking them clear to your reader. For example:

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 $$y_3y_4 = 3$$


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