Answer:
Attachment 1 : Graph B
Attachment 2 : Option B
Explanation:
( 1 ) The equation x = t² - 3 is represented by exponential growth, ( t² ) so it's graph will be similar to the first graph, graph 1, in our options. Then again we have to consider the equation y = √t - 2, which will be similar to graph 4, but with a greater slope. This leaves us with a solution of graph b.
( 2 ) We have the following system of equations at hand here.
{ x = 5 cot(t), y = - 3csc(t) + 4 }
Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,
x = 5 cot(t) ⇒ x - 5 = cot(t),
y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)
Now let's square these two equations, adding them --- Step #2
We know that csc²θ - cot²θ = 1, so let's subtract the equations
( y - 4 / - 3 )² = (csc(t))²
- ( x - 5 / 1 )² = (cot(t))²
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(y - 4)² / 9 - x² / 25 = 1
And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.