Question

Eliminate the parameter.

x = t2 + 1, y = t2 - 1

a.y = x + 2, x ≥ 1

b.y = x - 2, x ≥ 1

c.y = x2 - 2, x ≥ 1

d. y = x2 + 2, x ≥ 1

Answer 1

\[\begin{cases}x=t^2+1\\y=t^2-1\end{cases}\]
Since both \(x\) and \(y\) contain a \(t^2\) term, it's easy to solve for \(t^2\):
\[\begin{cases}t^2=x-1\\t^2=y+1\end{cases}\]
So, you have
\[x-1=y+1\\
~~~~~~~~~~\vdots\]

Answer 2

How would I simplify that into a standard form? @sinthsandgiggles

Answer 3

@SithsAndGiggles

Answer 4

Solve for \(y\) in terms of \(x\). Those are the provided answer choices, after all.

Answer 5

Do u knw differentintion......?

Answer 6

not at all @ashwinjohn3

Answer 7

Uhm, no differentiation required.

@grabyouroar, solve for \(y\) in the last equation I posted.
You get
\[y=x-2\]

Answer 8

OHH okay now I understand. I was mistaken for something else. Thanks!@SithsAndGiggles

Answer 9

You're welcome!

Answer 10

Would you mind helping me with one more question? @SithsAndGiggles

Answer 11

Ask away

Answer 12

Eliminate the parameter.

x = 7t, y = t + 4

Options are:
y = 7x + 4

y = 7x - 4

y =( x/7) + 4

y = (x/7) - 4
@SithsAndGiggles

Answer 13

This time, solving for \(t\) makes elimination simpler:
\[x=7t~\Rightarrow~t=\frac{x}{7}\\
y=t+4~\Rightarrow~t=y-4\]
So you're left with
\[\frac{x}{7}=y-4\]

Answer 14

So my end product will be a fraction?
@SithsAndGiggles

Answer 15

WAIT nvm lol

Answer 16

Solving for \(y\), you're left with
\[y=\frac{x}{7}+4\]
(the third answer choice)

The answer itself isn't really a fraction (more of a mixed number).

Answer 17

okay *face palm how do you know all this!?

Answer 18

Paying attention in class? Idk, this is pretty basic algebra haha

Answer 19

math's not my thing man! haha thanks again @SithsAndGiggles