Given the equations:
[tex]\begin{gathered} x=4t \\ \\ y=t^2-1 \end{gathered}[/tex]Let's identify the parametric equations which represent the same graph as the given equations.
Here, we have:
Rewrite the first equation for t
[tex]t=\frac{x}{4}[/tex]Now, plug in x/4 for t in the second equation:
[tex]\begin{gathered} y=(\frac{x}{4})^2-1 \\ \\ y=\frac{x^2}{4^2}-1 \\ \\ y=\frac{x^2}{16}-1 \end{gathered}[/tex]Also we have:
[tex]t^2=\frac{x^2}{16}[/tex]Rewrite equation 2 for t²
[tex]t^2=y+1[/tex]Hence, we have:
[tex]\begin{gathered} \frac{x^2}{16}=t^2=y+1 \\ \\ \frac{x^2}{16}=y+1 \\ \\ y=\frac{x^2}{16}-1 \end{gathered}[/tex]Now, let's solve for the following to confirm:
[tex]\begin{gathered} \text{ When x = 3t} \\ Substitute\text{ 3t for x:} \\ y=\frac{(3t)^2}{16}-1 \\ \\ y=\frac{9t^2}{16}-1 \\ \end{gathered}[/tex]Option B:
[tex]\begin{gathered} \text{ When x = 2t:} \\ \text{ Substitute 2t for x} \\ y=\frac{(2t)^2}{16}-1 \\ \\ y=\frac{4t^2}{16}-1 \\ \\ y=\frac{t^2}{4}-1 \\ \\ Option\text{ B is correct.} \end{gathered}[/tex]Option C:
[tex]\begin{gathered} \text{ When x =8t} \\ \text{ Substitute 8t for x:} \\ y=\frac{(8t)^2}{16}-1 \\ \\ y=\frac{64t^2}{16}-1 \\ \\ y=4t^2-1 \\ \\ Option\text{ C is correct.} \end{gathered}[/tex]Therefore, the correct parametric equations are:
[tex]\begin{gathered} B.\text{ x = 2t} \\ y=\frac{t^2}{4}-1 \\ \\ \\ \\ C.x=8t \\ y=4t^2-1 \end{gathered}[/tex]ANSWER:
• B. x = 2t
y = t²/4 - 1
• C. x = 8t
y = 4t² - 1