# Parametric to implicit form of a curve

"Find the implicit form of the curve defined by parametric equations $x = t+1,y=\frac{1}{t^{2}}$"

How can I clear $t$ to arrive at the implicit equation?

HINT:

If $x=\dfrac1{t+1}\iff t+1=\dfrac1x\iff t=\cdots$

If $x=t+1\iff t=x-1$

Put this value in $y=\dfrac1{t^2}$

Alternatively, $y=\dfrac1{t^2}\iff t^2=\dfrac1y\ \ \ \ (1)$

and $x=t+1\iff t=x-1\implies t^2=(x-1)^2\ \ \ \ (2)$

Compare the values of $t^2$ in $(1),(2)$ to eliminate $t$

• Sorry, I didn't put the right value for x, edited! – Nicolás Siplis Jul 1 '15 at 16:32
• @NicolásSiplis, It's even simpler! $t=x-1$ – lab bhattacharjee Jul 1 '15 at 16:33
• Wow, I just realized I was overcomplicating the solution by trying to clear $t$ from both $x$ and $y$. Thank you! – Nicolás Siplis Jul 1 '15 at 16:35
• @NicolásSiplis, Please find the alternative method – lab bhattacharjee Jul 1 '15 at 16:40
• I understand how you arrive at the result for both $x$ and $y$. However, I'm not sure if there's anything else I should do to get the implicit form from: $f(x) = 1/(x-1)^2$. I'm assuming all that's left is $1/(x-1)^2 - y = 0$, right? – Nicolás Siplis Jul 1 '15 at 16:43

Starting from $$x=t+1 \quad y=\frac{1}{t^2}$$ we get $$y = \frac{1}{(x-1)^2}$$ and thus $$F(x,y) = \frac{1}{(x-1)^2} - y = 0$$ as implicit curve equation.

Another approach:

$$y=\frac1{t^2}=\frac1{(t+1)^2-(2t+1)}=\frac1{x^2-(2x-1)}=\frac1{(x-1)^2}$$

and we're done.

An approach: express $t$ as a function of the other variables.

$$t=x-1=\pm\frac1{\sqrt y}.$$

In this case, it is better to express $t^2$ to avoid the double signs:

$$t^2=(x-1)^2=\frac1y.$$

Alternatively, say: "take $t+1$, subtract $1$ to get $t$, square and take the inverse to get $\dfrac1{t^2}$."

Then $\dfrac1{(x-1)^2}=y$.