hyperbola and circle (1 Viewer)

[juantheron]

Let be the only point to satisfying the hyperbola and


If and . then

options


1,2,4,none

 

[umm what]

Bro is it hsc math extension 2?

let be the only point to satisfying the hyperbola and


if and . Then

options


1,2,4,none

 

[jyu]

Let be the only point to satisfying the hyperbola and


If and . then

options


1,2,4,none

1

 

[juantheron]

jyu how can you get the answer

would you like to explain it to me

 

[jyu]

Let be the only point to satisfying the hyperbola and


If and . then

options


1,2,4,none

,

using given info,
,

.: and the hyperbola must be tangential to the circle which requires

.:

 

[juantheron]

Thanks jyu

but I did not understand the line

.: and the hyperbola must be tangential to the circle which requires

.:

Thanks

 

[jyu]

try to sketch the two graphs meeting the constraints

 

[juantheron]

try to sketch the two graphs meeting the constraints

I gave got the point that if hyperbola is tangent then there is only point of intersection

but did not understand how can we get

Thanks

 

[jyu]

at the point (2 of them) of contact, y=x, the circle and the hyperbola are symmetrical about the line y=x, .: a=b

 

[juantheron]

at the point (2 of them) of contact, y=x, the circle and the hyperbola are symmetrical about the line y=x, .: a=b

Thanks for answer

but i did not understand if circle and Rectangular hyperbola are symmetrical about

then how can i get

and (ii) doubt is how we can say that represent Rectangular Hyperbola

 

[jyu]

Thanks for answer

but i did not understand if circle and Rectangular hyperbola are symmetrical about

then how can i get

and (ii) doubt is how we can say that represent Rectangular Hyperbola

Sorry that I misled you. I wonder why my mistakes were not pointed out.

d^2=2a/(a+b) is ok. Points 0f contact of the 2 curves are at x=y=+/-sqrt(a/(a+b)), and gradient =-1.
Hence b=0 and d=sqrt2.

 

[juantheron]

Sorry that I misled you. I wonder why my mistakes were not pointed out.

d^2=2a/(a+b) is ok. Points 0f contact of the 2 curves are at x=y=+/-sqrt(a/(a+b)), and gradient =-1.
Hence b=0 and d=sqrt2.

Yes I didnot understand how you get a = b, Now I understand that things

Thanks jyu.