Question

Eliminate the parameter.

x = t2 + 2, y = t2 - 4

Answer 1

Just subtract the two equations and t^2 will get eliminated.

Answer 2

\(\bf x=t^2+2\implies x-2=t^2\implies \sqrt{x-2}={\color{blue}{ t}}
\\ \quad \\
y={\color{blue}{ t}}^2-4\implies y=?\)

Answer 3

y = x - 6, x ≥ 1? @jdoe0001

Answer 4

notice above... use substitution

Answer 5

oh the x is squared?

Answer 6

y = x2 - 6

Answer 7

hmmm ohh shoot.. ahemm you're correct dohh .. yes

Answer 8

wait i was right before? it would be y = x - 6, x ≥ 1?

Answer 9

\(\bf x=t^2+2\implies x-2=t^2\implies \sqrt{x-2}={\color{blue}{ t}}
\\ \quad \\
y={\color{blue}{ t}}^2-4\implies y=(\sqrt{x-2})^2-4\to x-2-4\to x-6\)

so long \(\bf x > 1\)

Answer 10

so yay i was right(: thanks

Answer 11

the domain constraint will apply to the rooted version thus
if x is equals to 1, that gives \(\bf \sqrt{1-2}\to \sqrt{-1}\) but if x is greater than 1, the radicand will be 0 or a positive value

Answer 12

so i was right..

Answer 13

yes, just keep in mind the domain is x > 1