one more parametric equation problem

Hello, scottydoint!

Quote:

$\begin{array}{cccc}{\color{blue}[1]} &x&=& \dfrac{1}{\sqrt{t-5}} \\ \\[-4mm] {\color{blue}[2]} & y&=& \dfrac{t}{t-5} \end{array}\quad\text{for }t \in (5,\,\infty)$

Square [1]: . $x^2 \:=\:\frac{1}{t-5} \quad\Rightarrow\quad t \:=\:\frac{1}{x^2} + 5$

Substitute into [2]: . $y \:=\:\frac{\frac{1}{x^2}+5}{\left(\frac{1}{x^2} + 5\right) - 5} \;=\;\frac{\frac{1}{x^2} + 5}{\frac{1}{x^2}}$

Multiply by $\frac{x^2}{x^2}\!:\quad y \:=\:\frac{x^2\left(\frac{1}{x^2} + 5\right)}{x^2\left(\frac{1}{x^2}\right)} \quad\Rightarrow\quad y \;=\;1 + 5x^2$

If $t \in(5,\infty)$ , then: . $\begin{array}{ccc}x \:\in\: (0,\infty) & \text{domain}\\ y \:\in\: (1,\infty) & \text{range}\end{array}$

The graph is an up-opening parabola, vertex at (0, 1) ... only the right half.

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