I think I am just overcomplicating things or am I going in the right direction? (pre-calculus)

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ScrollChamp

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There is a question about parametrics; x=t+3/t and y=t-3/t
I found the cartesian equation by rearranging the x equation into the quadratic form and solving for t=(x+/-sqrt(x²+12))/2 which I then substituted into y for y = quadratic - 3/quadratic.

Then there is the question to find the equation for the normal at point (4,-2). The abundance of +/-'s started to worry me and making me second guess myself. Is the question itself complicated or did I do it wrong?
 
Given that x= t+ 3/t and y= t- 3/t, (x, y)= (4, -2) we have t+ 3/t= 4 so t^2+ 3= 4t. t^2- 4t+ 3= (t- 1)(t- 3)= 0. If t= 1, y= 1- 3= -2 and when x= 3, y= 1- 3/3= 0. So t must be 1 at this point.

dx/dt= 1- 3/t^2, which is 1- 3= -2 at (4, 2), and dy/dt= 1+ 3/t^2, which is 1+ 3= 4 at (4, 2).

The slope of the tangent line at (4, 2) is dy/dx= (dy/dt)/(dx/dt)= 4/-2= -2.

The normal line has slope -1/(-2)= 1/2. The line through (4, 2) with slope 1/2 is y= (1/2)(x- 4)+ 2.
We can write that as 2y= x- 4+ 4= x or 2y= x, y= x/2.
 
ScrollChamp said:
There is a question about parametrics; x=t+3/t and y=t-3/t
I found the cartesian equation by rearranging the x equation into the quadratic form and solving for t=(x+/-sqrt(x²+12))/2 which I then substituted into y for y = quadratic - 3/quadratic.

Then there is the question to find the equation for the normal at point (4,-2). The abundance of +/-'s started to worry me and making me second guess myself. Is the question itself complicated or did I do it wrong?
Click to expand...
Please state the entire problem as given to you. That may make a big difference.

In particular, were you asked for the Cartesian equation, or did you do that just because you thought it was necessary? If you were only asked for the normal line, it isn't.
 
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HallsofIvy said:
The normal line has slope -1/(-2)= 1/2. The line through (4, 2) with slope 1/2 is y= (1/2)(x- 4)+ 2.
We can write that as 2y= x- 4+ 4= x or 2y= x, y= x/2.
Click to expand...
Note - the point should be (4,-2)
 
Your're welcome. All you want is to get some sort of equation connecting x and y and with no t. In doing that, you can do anything that's mathematically legal!
 
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