Question 1073403
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Circle T intersects the hyperbola y=1/x at (1,1), (3, 1/3), and two other points. 
What is the product of the y coordinates of the other two points? Please write in proof format. Thank you.
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<pre>
The equation of the circle is

{{{(x-a)^2 + (y-b)^2}}} = {{{r^2}}},

for some "a", "b" and "r".

The equation of the hyperbola is y = {{{1/x}}} (given !)

You will get the equation for common points (intersection points) if you substitute equation (2)  into the equation (1).
You will get

{{{(x-a)^2}}} + {{{(1/x - b)^2}}} = {{{r^2}}},   or

{{{x^2 -2ax + a^2}}} + {{{1/x^2}}} - {{{(2b/x)}}} + {{{b^2 - r^2}}} = 0.

Next multiply both sides by {{{x^2}}} to rid of denominators. You will get

{{{x^4 - 2ax^3 +a^2*x^2 + 1 - 2bx + b^2*x^2 -r^2*x^2}}} = 0,   or, ordering by descending degrees of x

{{{x^4 - 2a*x^3 + (a^2 + b^2 - r^2)*x^2 - 2bx + 1}}} = 0.


The last equation is the 4-th degree equation. Its roots are x-coordinates of the common (intersection) points.

Two of the roots are given: they are x-coordinates of the given intersection points  x= 1 and x= 3.

Two other roots are not known.
But, according to the Vieta's theorem for the equation of the degree 4, the product of four roots is the constant term  

           ( <U>! - it is the KEY idea ! </U>).


Thus, {{{x[1]*x[2]*x[3]*x[4]}}} = {{{1*(1/3)*x[3]*x[4]}}} = 1,      (1)

which implies

      {{{x[3]*x[4]}}} = {{{1/((1/3))}}} = 3.               (2)

The problem asks about {{{y[3]*y[4]}}}, but it is simply 

     {{{y[3]*y[4]}}} = {{{1/x[3]}}}.{{{1/x[4]}}} = {{{1/(x[3]*x[4])}}} = {{{1/3}}}

due to (2).


So, the problem is solved and the answer is: the product of y-coordinates of the two other intersection points is {{{1/3}}}.


<U>Answer</U>.  The product of y-coordinates of the two other intersection points is {{{1/3}}}.
</pre>

Solved.



For Vieta's Theorem see <A HREF=https://en.wikipedia.org/wiki/Vieta's_formulas>this Wikipedia article.