Question 373977
Which of the following is an equation of one of the asymptotes of the hyperbola:


{{{3x^2-3y^2=48}}}

First we must get the equation in either the form {{{x^2/a^2-y^2/b^2=1}}} or
{{{y^2/a^2-x^2/b^2=1}}}.  Either way we get a 1 on the right side by dividing every term
by 48.

{{{3x^2/48-3y^2/48=48/48}}}

{{{x^2/16-y^2/16=1}}}

Comparing to {{{x^2/a^2-y^2/b^2=1}}}, we see that 
aČ=16 and bČ=16 and so a=4, b=4

The defining rectangle is then:

{{{drawing(400,400,-10,10,-10,10, 

graph(400,400,-10,10,-10,10), green(rectangle(-4,-4,4,4),line(-4,4,4,4),
line(-4,-4,4,-4)

) )}}}

So its extended diagonals are the asymptotes:

{{{drawing(400,400,-10,10,-10,10, 

graph(400,400,-10,10,-10,10,-sqrt(x^2-16)), green(rectangle(-4,-4,4,4),line(-4,4,4,4),
line(-4,-4,4,-4), line(-11,-11,11,11),line(-11,11,11,-11),

graph(400,400,-10,10,-10,10,sqrt(x^2-16))  

) )}}}

And they have slope ±1 and pass through the origin so their equations are
y = x  and y = -x

In standard form these are

x-y=0 and x+y=0

x-y=0 is one of the choices.  But I notice that there is also
the choice 3x-3y=0, which is equivalent to x-y=0.  So either
of these would technically be correct.

Edwin</pre>