Q on hyperbolic equation

[Sonal7]

The question is about a rectangular hyperbola xy=c ²
The general point (ct, [MATH]\frac{c}{t}[/MATH])
What is the value of t at point (-2,8)? This is the first part and this confused me.
I am unsure how to find t by rearranging ct and [MATH]\frac{c}{t}[/MATH], the equating the two expression for t.
my working is as below:
ct = -2
hence c=[MATH]\frac{-2}{t}[/MATH][MATH]\frac{c}{t}[/MATH]=8
hence c=8t
equating the two equations:
[MATH]\frac{-2}{t}=8t[/MATH]t²=-1/4
now what do i do? I know the answer is wrong as the question wants you to give the answer as a whole number of a fraction expressed as p/q.

EDIT: I have the correct answer, now. It turns out that the co-ordinate is meant to be (-2,-8) in the solutions. I would have been right if they hadn't made so many mistakes.

 

[Romsek]

[MATH] c t = -2\\ \dfrac c t = 8\\ c = 8t\\ 8t^2 = -2\\ t = \pm \dfrac i 2\\ c = \pm 4i [/MATH]
Are you sure there was a negative sign?

 

[Sonal7]

Yes but I graphed the xy when either is negative and this causes a problem, as c² is suppose to be positive. I am doing some work around this. I dont think we are expected to give an imaginary number as an answer.

 

[Dr.Peterson]

Please state the entire problem as given to you. This includes everything it says about the expected form of the answer, and conditions such as whether c is assumed to be real.

It could just be a mistake on their part, but we can't say that without seeing the entire thing (ideally as an image).

 

[Sonal7]

Please note the answers have been amended, it not how i entered them. I am working through these the second time!

 

[Sonal7]

I got the first bit right.

I just thought of it graphically and I thought it might be -1/2 based on the quadrant the curve is in. I think as c is meant to be positive, then t must be -1/2.
I am now stuck on (c). I cannot figure out where I am going wrong.

This part is a good question, the quation is about finding the equation of the normal at (2x, 1/2c). This immediately gives us the value of t as
we know y is [MATH]\frac{c}{t}[/MATH]Now we have
[MATH]ct=2x[/MATH][MATH]2c=2x[/MATH]This means x =c

the derivate of y is -c²/x²
so the gradient of normal is x²/c²
so now we have
y-1/2 c= x²/c²(x-2x)
hence y-1/2c =1 (x-2x)
this give y =-x+1/2c
its the wrong answer! Please help.

 

[Sonal7]

[MATH] c t = -2\\ \dfrac c t = 8\\ c = 8t\\ 8t^2 = -2\\ t = \pm \dfrac i 2\\ c = \pm 4i [/MATH]
Are you sure there was a negative sign?

Yes it turns out that the co-ordinate is meant to be (-2,-8) in the solutions. I had to pass the test to get to the solutions which i did in the end. I would have been right if they hadnt made so many mistakes. I am going to close this discussion so that people dont waste their time spotting errors.