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# Algebra Problem

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

Guest Mar 17, 2017
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### 1+0 Answers

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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

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