Find the vertex of the transformed function

Full solution

13

. A curve

C

has parametric equations

\newline

x=\frac{t^{2}+5}{t^{2}+1} \quad y=\frac{4 t}{t^{2}+1}

\newline

Show that all points on

C

satisfy

\newline

(x-3)^{2}+y^{2}=4

Calculate $x - 3$ : We are given the parametric equations for the curve $C$ : $x = \frac{t^2 + 5}{t^2 + 1}$ $y = \frac{4t}{t^2 + 1}$ We need to show that these satisfy the equation $(x - 3)^2 + y^2 = 4$ .
Calculate y: First, let's calculate $(x - 3)^2$ using the given expression for x: $\newline$ $x - 3 = \frac{(t^2 + 5)}{(t^2 + 1)} - 3$ $\newline$ $x - 3 = \frac{(t^2 + 5) - 3(t^2 + 1)}{(t^2 + 1)}$ $\newline$ $x - 3 = \frac{(t^2 + 5) - (3t^2 + 3)}{(t^2 + 1)}$ $\newline$ $x - 3 = \frac{-2t^2 + 2}{(t^2 + 1)}$ $\newline$ Now, we square $(x - 3)$ : $\newline$ $(x - 3)^2 = \frac{(-2t^2 + 2)^2}{(t^2 + 1)^2}$
Combine $x - 3$ and $y$ : Next, we calculate $y^2$ using the given expression for $y$ :
$y = \frac{4t}{t^2 + 1}$
$y^2 = \left[\frac{(4t)^2}{(t^2 + 1)^2}\right]$
$y^2 = \frac{16t^2}{(t^2 + 1)^2}$
Simplify numerator: Now, we add $(x - 3)^2$ and $y^2$ :
$(x - 3)^2 + y^2 = [(-2t^2 + 2)^2] / [(t^2 + 1)^2] + [16t^2] / [(t^2 + 1)^2]$
Since both terms have the same denominator, we can combine them:
$(x - 3)^2 + y^2 = [(-2t^2 + 2)^2 + 16t^2] / [(t^2 + 1)^2]$
Factorize numerator: We simplify the numerator: $\newline$ $(-2t^2 + 2)^2 + 16t^2 = (4t^4 - 8t^2 + 4) + 16t^2$ $\newline$ $(-2t^2 + 2)^2 + 16t^2 = 4t^4 + 8t^2 + 4$
Substitute back: We notice that the numerator can be factored: $\newline$ $4t^4 + 8t^2 + 4 = 4(t^4 + 2t^2 + 1)$ $\newline$ Recognizing that $t^4 + 2t^2 + 1$ is a perfect square, we can write it as: $\newline$ $4(t^4 + 2t^2 + 1) = 4(t^2 + 1)^2$
Cancel out terms: Now, we substitute this back into our equation for $(x - 3)^2 + y^2$ : $(x - 3)^2 + y^2 = \frac{4(t^2 + 1)^2}{(t^2 + 1)^2}$ We can cancel out the $(t^2 + 1)^2$ terms: $(x - 3)^2 + y^2 = 4$
Final conclusion: We have shown that for any value of $t$ , the equation $(x - 3)^2 + y^2 = 4$ holds true for the given parametric equations of the curve $C$ . Therefore, all points on $C$ satisfy the equation.