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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?

Guest Oct 9, 2018
 #1
avatar+3190 
+2

I'm not seeing any simple way of doing this though the answer is simple enough it makes me suspect there is one.

 

Rom  Oct 10, 2018
 #2
avatar+94110 
+1

Idk either 

 

I found that the centre of the circle is    \((a,\frac{9a-16}{3})\)      where a is int the set ot reals

 

and the radius sqared iis                   \(r^2= 10a^2-40a+\frac{370}{9}\)

 

so the equation of the circle is

 

\((x-a)^2+(y-\frac{9a-16}{3})^2= 10a^2-40a+\frac{370}{9}\)

 

I know that is right becasue I have graphed it but when i take it further I get in a mess.  :/

Melody  Oct 10, 2018
 #3
avatar+27229 
+3

Here's my take:

 

.

Alan  Oct 10, 2018
 #4
avatar
+2

It isn't necessary to calculate the equation of the circle, nor even the co-ordinates of its centre.

Melody calculated the co-ordinates of the centre as (a, (9a - 16)/3), but to save on the typing I'll use (a, A).

 

All we need do is to find two other points on the hyperbola that are the same distance from the centre as the two given points.

 

Let the co-ordinates of a general point on the hyperbola be (p, 1/p), then we need, (using the point (1, 1) as one of the given points),

(p - a)^2 + (1/p - A)^2 = (1 - a)^2 + (1 - A)^2,

multiplying that out,

p^2 + a^2 -2ap + (1/p)^2 + A^2 - 2A/p = 1 + a^2 -2a + 1 + A^2 - 2A,

cancelling the a^2 and the A^2 and then multiplying throughout by p^2 gets you

p^4 - 2ap^3 + 2(a + A - 1)p^2 - 2Ap + 1 = 0.

Suppose that the roots of this are r, s, t, u, so that 

(p - r)(p - s)(p - t)(p - u) = 0,

then, after expanding (which you don't need to do !) and equating constants with the actual equation, rstu = 1.

However, two of the  roots are known, they're 1 and 3, so the product of the other two must be 1/3.

 

That means that the product of the other two y co-ordinates will be 3. 

 

Tiggsy

 

(Sorry, I couldn't get LaTeX to work for some reason.)

Guest Oct 10, 2018
edited by Guest  Oct 10, 2018
 #5
avatar+92759 
+1

Nice, Alan  and Tiggsy  !!!

 

 

cool cool cool

CPhill  Oct 10, 2018
 #6
avatar+94110 
+1

Yes, thanks guys :))

Melody  Oct 10, 2018

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