Question 252347
If P is the any point on the hyperbola whose axis are equal,prove that {{{SP*S'P=CP^2}}}, where S and S' are the foci, and C is the center.
  
<pre><font size = 4 color = "indigo"><b>
Let the tranverse axis be along the x-axis and the conjugate axis be
along the y-axis, with the center C at (0,0), the origin.

Let both axes be 2, so that both the semi-tranverse axis, "a", and 
semi-conjugate axis, "b", are 1 each.   Then the equation of the 
hyperbola, which is {{{x^2/a^2-y^2/b^2=1}}}, becomes simply {{{x^2-y^2=1}}}.
In a hyperbola, {{{c^2=a^2+b^2}}}, so {{{c^2=1^2+1^2=1+1=2}}}, therefore
{{{c = sqrt(2)}}}, where "c" is the distance from the center to the focus.
Therefore S and S' are the points ({{{""+-sqrt(2)}}},0).  

[Do not confuse the center "C(0,0)" with the value of "c", the distance 
from the center to each focus.]

Let P(x,y) be any arbitrary point on the hyperbola:
The blue line is CP:

The graph is:

{{{drawing(400,400,-3,3,-3,3, rectangle(-1,-1,1,1),
graph(400,400,-3,3,-3,3,sqrt(x^2-1),x), blue(line(0,0,1.7,1.374772708)),
graph(400,400,-3,3,-3,3,-sqrt(x^2-1),-x),
line(1.7,1.374772708, sqrt(2),0), line(1.7,1.374772708, -sqrt(2),0),
locate(1.8,1.4,"P(x,y)"), locate(1.5,.35,S(sqrt(2),0)), l
ocate(-2.31,.35,"'"), locate(-2.35,.36,S(-sqrt(2),0)) 
 )}}} 

Using the distance formula to find SP and SP' in terms of x:

{{{SP = sqrt((x-sqrt(2))^2+(y-0)^2) = sqrt(x^2-2x*sqrt(2)+2+y^2)}}}

Since the equation of the hyperbola is {{{x^2-y^2=1}}}, then {{{y^2=x^2-1}}},
so substituting we get:

{{{SP = sqrt(x^2-2x*sqrt(2)+2+y^2)=SP = sqrt(x^2-2x*sqrt(2)+2+(x^2-1))}}}

{{{SP = sqrt(x^2-2x*sqrt(2)+2+x^2-1)= sqrt(2x^2-2x*sqrt(2)+1)}}}

Similarly,

{{{"SP'" = sqrt((x-(-sqrt(2)))^2+(y-0)^2) = sqrt((x+sqrt(2))^2+y^2)=
sqrt(x^2+2x*sqrt(2)+2+y^2)}}}

As before, since the equation of the hyperbola is {{{x^2-y^2=1}}},
then {{{y^2=x^2-1}}}, so substituting we get:

{{{"SP'" = sqrt(x^2+2x*sqrt(2)+2+y^2)=SP = sqrt(x^2+2x*sqrt(2)+2+(x^2-1))}}}

{{{"SP'" = sqrt(x^2+2x*sqrt(2)+2+x^2-1)= sqrt(2x^2+2x*sqrt(2)+1)}}}

So 

{{{SP*"SP'"= sqrt(2x^2-2x*sqrt(2)+1)*sqrt(2x^2+2x*sqrt(2)+1)}}}

{{{SP*"SP'"= sqrt((2x^2+1)-2x*sqrt(2))*sqrt((2x^2+1)+2x*sqrt(2))}}}
      
Multiplying under the radicals:

{{{SP*"SP'"= 
sqrt((2x^2+1)^2-(2x*sqrt(2))^2) =sqrt((4x^4+4x^2+1)-(4x^2*2))  =

sqrt(4x^4+4x^2+1-8x^2) }}}

{{{SP*"SP'"= sqrt(4x^4-4x^2+1) = sqrt((2x^2-1)^2) = abs(2x^2-1)  }}}

Next we use the distance formuls to find CP where C is the origin (0,0).

{{{CP = sqrt((x-0)^2+(y-0)^2)=sqrt(x^2+y^2)}}}

Since the equation of the hyperbola is {{{x^2-y^2=1}}}, then {{{y^2=x^2-1}}},
so substituting we get

{{{CP =sqrt(x^2+y^2)=sqrt(x^2+x^2-1)= sqrt(2x^2-1)}}}

so {{{CP^2 = (sqrt(2x^2-1))^2 = abs(2x^2-1)}}}

Therefore {{{SP*"SP'" = CP^2}}}, because both equal {{{abs(2x^2-1)}}} 

Edwin</pre>