Question

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Answer 1

Answer:

1. Parabola

2. Choice A

Step-by-step explanation:

1. We have been given the following parametric equations;

$x=t+3\\y=\frac{2t^{2} }{3}-2$

We first need to obtain the equation that connects x and y directly;

we solve for t in the first equation,

x = t + 3

t = x - 3

This implies that we shall substitute t with x - 3 in the second equation;

$y=\frac{2(x-3)^{2} }{3}-2\\y=\frac{2}{3}(x-3)^{2}-2$

The graph of the function is shown in the attachment below. The conic section created by the parametric equations is thus a parabola.

2.

In the next question, the following parametric equations have been given;

$x=\sqrt{16-t^{2} } \\y=t$

It is much easier to solve for t in the second equation;

t is simply equals to y; t = y

The next step is to substitute t with y in the first equation;

$x=\sqrt{16-y^{2} }$

We then square both sides to eliminate the square root;

$x^{2} =16-y^{2}\\x^{2}+y^{2}=16$

Thus, choice A is the correct answer.