1. Parametric equation

Teh graph of xy-4x-2y-4=0 can be expressed as a set of parametric equations. If y = 4t/(t-3) and x= f(t), then f(t)=? The answer is t-1. Thanks for advice for tackling these in general.

2. Re: Parametric equation

Why not just go ahead and make the substitution?

$\displaystyle xy-4x-2y-4=0$

$\displaystyle [f(t)]\frac{4t}{t-3}-4[f(t)]-2\cdot\frac{4t}{t-3}-4=0$

Solve for $\displaystyle [f(t)]$

3. Re: Parametric equation

My simplication is getting really sloppy. I need some help with solving for x (f(t))

4. Re: Parametric equation

First, separate the function, with terms including $\displaystyle f(t)$ on the left side of the equation and terms without an $\displaystyle f(t)$ component on the right. Then, take out a common factor of $\displaystyle f(t)$ from the left. Write everything remaining on the left as one fraction, and everything on the right as one fraction. Then divide/multiply through as necessary so that you're left with something of the form $\displaystyle f(t)=\cdots$ to simplify. If you need further help, please try this method yourself and show me how far you can get.

5. Re: Parametric equation

Originally Posted by Quacky
Why not just go ahead and make the substitution?

$\displaystyle xy-4x-2y-4=0$

$\displaystyle [f(t)]\frac{4t}{t-3}-4[f(t)]-2\cdot\frac{4t}{t-3}-4=0$

Solve for $\displaystyle [f(t)]$
f(t)4t/t-3 - 4f(t) = 4(t-3)/-8t
f(t) (4t/t-3 - 4)

I then added the numbers between the parenthesis by multipling -4 by t-3. Then I multipled/divided accordingly to get it to the other side. I came up with t^(2)-6t+9/(-32t^(2) + 48)

6. Re: Parametric equation

Originally Posted by Quacky
Why not just go ahead and make the substitution?

$\displaystyle xy-4x-2y-4=0$

$\displaystyle [f(t)]\frac{4t}{t-3}-4[f(t)]-2\cdot\frac{4t}{t-3}-4=0$

Solve for $\displaystyle [f(t)]$
Originally Posted by benny92000
f(t)4t/t-3 - 4f(t) = 4(t-3)/-8t
f(t) (4t/t-3 - 4)

I then added the numbers between the parenthesis by multipling -4 by t-3. Then I multipled/divided accordingly to get it to the other side. I came up with t^(2)-6t+9/(-32t^(2) + 48)
I still feel like you're not putting a full commitment into it.

$\displaystyle [f(t)]\frac{4t}{t-3}-4[f(t)]-2\cdot\frac{4t}{t-3}-4=0$

I then advised that you split the terms so that terms involving an $\displaystyle f(t)$ were on the left and terms without were on the right, and take a common factor of $\displaystyle f(t)$

$\displaystyle [f(t)](\frac{4t}{t-3}-4)=4+\frac{8t}{t-3}$

My next piece of advice was to rewrite everything on each side over a common denominator.

$\displaystyle [f(t)]\cdot\frac{4t-4(t-3)}{t-3}=\frac{4(t-3)+8t}{t-3}$

Now we can just multiply through by $\displaystyle (t-3)$, as my previous response implied, to get:

$\displaystyle [f(t)]\cdot [4t-4(t-3)]=[4(t-3)+8t]$

I've done the majority for you now - see if you can finish.

7. Re: Parametric equation

I fnally came to the right answer. Thanks