Respuesta :

To find:

The point which are on the plane curve.

Solution:

Given parametric equations are x = 2t + 5 and y = 3t^2.

Find the value of t in terms of x.

[tex]\begin{gathered} x=2t+5 \\ x-5=2t \\ \frac{x-5}{2}=t \end{gathered}[/tex]

substitute this value of t in y:

[tex]y=3(\frac{x-5}{2})^2[/tex]

Now, check if every point passes through curve.

at x = 1,

[tex]\begin{gathered} y=3(\frac{1-5}{2})^2 \\ y=3(-2)^2 \\ y=12 \end{gathered}[/tex]

So, the curve does not pass through (1, 7).

at x = 2,

[tex]\begin{gathered} y=3(\frac{2-5}{2})^2 \\ y=3(\frac{9}{4}) \\ y=\frac{27}{4} \end{gathered}[/tex]

So, the curve does not pass through (2, 9).

at x= 3,

[tex]\begin{gathered} y=3(\frac{3-5}{2})^2 \\ y=3(1)^2 \\ y=3 \end{gathered}[/tex]

so, the curve does not pass through (3, -3).

at x = 3, y = 3, So, the curve passes through (3, 3).

at x = 7.

[tex]\begin{gathered} y=3(\frac{7-5}{2})^2 \\ y=3(1)^2 \\ y=3 \end{gathered}[/tex]

So, the curve passes through (7, 3).

Thus, options D and E are correct.

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