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Susan Munoz

Answered question

2021-12-03

Find two different sets of parametric equations for the rectangular equation

y = 4 x + 1, t = - 1

at the point

(- 2, - 7)

Answer & Explanation

George Spencer

Beginner2021-12-04Added 12 answers

Step 1
Given equation is $y = 4 x + 1$
Consider parametric equation for x to be $x = t + b$ where b is constant.
To find value of b substitute $x = - 2$ and $t = - 1$ in $x = t + b$ ,
$\Rightarrow - 2 = - 1 + b$
$\Rightarrow b = - 1$
So $x = t + b$ becomes $x = t - 1$
Now substitute $x = t - 1$ in $y = 4 x + 1$
$y = 4 (t - 1) + 1$
$\Rightarrow y = 4 t - 4 + 1$
$\Rightarrow y = 4 t - 3$
So the first solution is
${\begin{cases} x = t - 1 \\ y = 4 t - 3 \end{cases}$
Step 3
To find another solution:
Consider parametric equation for x to be $x = t$
Now to find y substitute $x = t$ in $y = 4 x + 1$ we get,
$\Rightarrow 4 = 4 t + 1$
So the second solution is
${\begin{cases} x = t \\ y = 4 t + 1 \end{cases}$
Therefore two different sets of parametric equations are
${\begin{cases} x = t - 1 \\ y = 4 t - 3 \end{cases}$
and
${\begin{cases} x = t \\ y = 4 t + 1 \end{cases}$

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