Conics - Rectangular Hyperbola (1 Viewer)

RivalryofTroll

Sleep Deprived Entity
Joined
Feb 10, 2011
Messages
3,805
Gender
Male
HSC
2013
Uni Grad
2019
Prove that the hyperbola with equation x^2 - y^2 = a^2 is the hyperbola xy = 1/2.a^2 referred to different axes.

Do you use perpendicular distances or something?

Not sure how to do it.
 

qrpw

Member
Joined
Apr 8, 2010
Messages
83
Gender
Male
HSC
2012
Take a random point (x1, y1) on the hyperbola, and multiply it by (cis 45) to rotate it anticlockwise. So something like (x1 + iy1)(cis45). Then solve the locus when you get your new x and y-coordinates.
 

RivalryofTroll

Sleep Deprived Entity
Joined
Feb 10, 2011
Messages
3,805
Gender
Male
HSC
2013
Uni Grad
2019
Take a random point (x1, y1) on the hyperbola, and multiply it by (cis 45) to rotate it anticlockwise. So something like (x1 + iy1)(cis45). Then solve the locus when you get your new x and y-coordinates.
Is there a way to do it without Complex Numbers?
 

FdashX

New Member
Joined
Feb 12, 2013
Messages
7
Gender
Undisclosed
HSC
2013
when theres a way of solving with complex numberz
ALWAYS used complex numberz!

complex numberZ ruleZ
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Look up Terry Lee - utilises perpendicular distances from tangents.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top