FP1 Rectangular Hyperbola question

Why is it called rectangular if it doesn't look like a rectangle and actually looks more like a reciprocal function?

y=(c^2)/x
y=16/x

Then c^2=16
c=+or -4. But c>0 so c=4
But WHY can't c=-4? Surely the square in c^2 sorts it out?

Thanks

A hyperbola with perpendicular asymptotes is called a rectangular hyperbola.

c cannot be -4 because, as you said yourself, c>0.

I still don't get why it is called a rectangular hyperbola

But the thing is y=(c^2)/x, whether I have c=-4 or c=4, we get the numerator of the RHS as 16. Hence with, say x=2, we end up with the same result of y=16/2=8 whether we use c=-4 or c=4. Hence when solving, say, 8=(c^2)/2, I ask WHY is c=-4 invalid as a root?

just accept that, by definition, a hyperbola with perpendicular asymptotes is called a "rectangular" hyperbola.
And c can be -4 unless the question has specified that it is equal to or greater than 0.

J

for a) it`s just a matter of using the equation of the given hyperbola with c>0, in the form:

$y=\frac{c^2}{x}$ and equating it to ANY line with equation $y=-mx, m>0$ trough the origin. this equation indicates...

for a) it`s just a matter of using the equation of the given hyperbola with c>0, in the form:

$y=\frac{c^2}{x}$ and equating it to ANY line with equation $y=-mx, m>0$ trough the origin. this equation indicates...