concis q (1 Viewer)

[onebytwo]

hello people,

just wanted to know for the equation 2xy - 3x - y - 2 = 0
- how would you rewrite it in hyperbolic form
- how would show its a hyperbola

thanks for any help

 

[_ShiFTy_]

I dont think that could be converted to the 4U Hyperbolic form...

2xy - 3x - y - 2 = 0
y(2x - 1) = 3x + 2
y = (3x + 2)/(2x - 1)

 

[sasquatch]

2xy - 3x - y - 2 = 0

2x(y - (3/2)) - y = 2
2x(y - (3/2)) - y + (3/2) = (7/2)
2x(y - (3/2)) -(y - (3/2)) = (7/2)
(2x - 1)(y - (3/2)) = (7/2)
(x - (1/2))(y - (3/2)) = (7/4)

From what i learnt from someone else on the forum, to show its a rectangular hyperbola, you can show that the asymptotes are at right angles to each other.

x = (1/2) and y = (3/2) are the assymptotes, which are perpendicular to each other, hence the hyperbola is rectangular.

 

[_ShiFTy_]

2xy - 3x - y - 2 = 0

2x(y - (3/2)) - y = 2
2x(y - (3/2)) - y + (3/2) = (7/2)
2x(y - (3/2)) -(y - (3/2)) = (7/2)
(2x - 1)(y - (3/2)) = (7/2)
(x - (1/2))(y - (3/2)) = (7/4)

From what i learnt from someone else on the forum, to show its a rectangular hyperbola, you can show that the asymptotes are at right angles to each other.

x = (1/2) and y = (3/2) are the assymptotes, which are perpendicular to each other, hence the hyperbola is rectangular.

Hmmm, forgot about the xy=c^2 form

 

[onebytwo]

i get that
also, what if they asked to find the foci and directices or whatever???
how would you go about that

 

[Riviet]

Correct if I'm wrong, but I think you would just use S(+ae,0) and x=+a/e just like in a standard hyperbola. Also remember that e=sqrt2 in this rectangular hyperbola.

 

[felixcthecat]

yupyup..in general
a^2 = (b^2)[(e^2)-1] to find focii and directrices given by riviet

for rectangular hyperbolas
xy=c^2

directrices x+y=+/- (sqrt2)c
focii [+/-(sqrt2)c, +/-(sqrt2)c]
vertices (+/- c, +/- c)