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 a proof form. Thank you. Hint: try to make a quadratic with the y-coordinates

Guest Mar 17, 2017

#1**0 **

I'm sure this could have been done much more easily than my approach......as it is..... it's messy.....but "do-able"

First of all.......we can draw a chord between (1,1) and (3,1/3)......

the midpoint of this chord is (2, 2/3)

Now a line drawn perpendicular to this chord will go through the center of the circle that we are seeking.......the slope of this line will be the negative reciprocal of ( 1 - 1/3) /(1 - 3) = (2/3)/-2 =

-1/3 ....so the slope is 3

And the equation of this line will be y = 3(x -2) + 2/3 → y = 3x - 6 + 2/3 → y = 3x -16/3

Now the intersection of this line and the hyperbola can be found as follows

1/x = 3x - 16/3

1 = 3x^2 - (16/3)x

3x^2 - (16/3)x - 1 = 0

Using the quadratic formula, the x coordinate of the intersection we are interested in will occur in the 3rd quadrant = [8 - sqrt (91)] / 9

And the y value of this point is the reciprocal of this = 9 /[ 8 - sqrt(91)]

So......to find the center of the circle (x,y) passing through this point and (1,1) and (3,1/3) we need to solve this:

(x - 1)^2 + (y - 1)^2 = (x -3)^2 + (y - 1/3)^2 = (x - [8 - sqrt (91)] / 9)^2 + (y - 9 /[ 8 - sqrt(91)])^2

This is obviously messy to solve so I will lean on Wolframalpha, here

The center is located at ( 2 - sqrt(91)/9, 2/3 - sqrt(91)/3 )

Now.....to find the radius between this point and (1,1) we have :

sqrt [ ( 2 - sqrt(91)/9 -1)^2 + ( 2/3 - sqrt(91)/3 - 1)^2

Again....with a little CAS help....the radius is 10sqrt(10)/9

So the equation of our circle is

(x - 2 + sqrt(91)/9)^2 + ( y - 2/3 + sqrt(91)/3) ^2 = (10sqrt(10)/ 9)^2

And the x coordinates of the intersection of this circle and y = 1/x can be found by substituting 1/x for y in the equation of the circle...again....a CAS is helpful here

So we have

(x - 2 + sqrt(91)/9)^2 + ( 1/x - 2/3 + sqrt(91)/3) ^2 = (10sqrt(10)/ 9)^2

One more time with a CAS, we have

The x intersection points are - (8 + sqrt(91))/ 9 and (8 - sqrt(91)) /9

And the y intersection points are the reciprocal of these....so we have.....

[-9 / (8 + sqrt(91)] and [9/ {8 - sqrt(91)]

And the product of these =

-81 / [ 64 - 91] =

-81 / -27 =

3

Here's the graph demonstrating some of this : https://www.desmos.com/calculator/5jxqoyoik4

CPhill
Mar 18, 2017