Conics Q (1 Viewer)

[azureus88]

The hyperbola x^2/a^2 - y^2/b^2 =1 has one focus S on the positive x axis. The circle with centre S and radius b cuts the hyperbola at points R and T. If b=a, show that RT is the diameter of the circle.

when i attempted this question i got:

(x-ae)^2 + y^2 = a^2
x^2 - y^2 = a^2

Solving simultaneously,

2x^2 - 2aex + (a^2e^2 - 2a^2) = 0

If midpoint of RT is (X,Y)

then X= (2ae/2)/2 = ae/2 which isnt the same x coordinate as S.

What have i done wrong?

 

[azureus88]

2x^2 - 2aex + (a^2e^2 - 2a^2) = 0

subbing in e=√2, i get

2x^2 - 2a√2x + (2a^2 - 2a^2) = 0
x^2 - a√2x = 0
x(x-a√2) = 0
x=0, a√2

so midpoint would be a√2/2 :S

 

[azureus88]

yeh i was thinking that too, but if you consider it using the locus definition of hyperbola, you get a different answer.

SR = ST = b (equal radii)

b=e(x- (a/e)) (SR=ST=ePM)
x=(a+b)/e = 2a/e = asqrt(2)

S(ae,0)=>(asqrt(2),0)

How come the original method didnt work out the same way?

 

[Trebla]

yeh i was thinking that too, but if you consider it using the locus definition of hyperbola, you get a different answer.

SR = ST = b (equal radii)

b=e(x- (a/e)) (SR=ST=ePM)
x=(a+b)/e = 2a/e = asqrt(2)

S(ae,0)=>(asqrt(2),0)

How come the original method didnt work out the same way?

It did work out in a similar way as you still get the solution x = a√2. I think we are meant to ignore x = 0 in the earlier method because it is not a feasible solution since subtitution into the second equation yields a false statement.

 

[azureus88]

oh ok then, thanks