Question

identify the conic given by $\frac{(x - 10)^2}{36}-\frac{(y - 9)^2}{36}=1$

ellipse

parabola

circle

hyperbola

question 5 (5 points)

write the parametric equations in rectangular form.

$y = 2t^2+6$

$x = 5t - 8$

$y=\frac{2}{25}x^2+\frac{32}{25}x+\frac{278}{25}$

$y=\frac{2}{25}x^2-\frac{32}{25}x+\frac{278}{25}$

$y=\frac{2}{25}x^2+\frac{32}{25}x+\frac{22}{25}$

$y=-\frac{2}{25}x^2+\frac{32}{25}x+\frac{278}{25}$

Explanation:

Step1: Solve \(x = 5t-8\) for \(t\)

From \(x = 5t - 8\), we get \(t=\frac{x + 8}{5}\)

Step2: Substitute \(t\) into \(y = 2t^{2}+6\)

Substitute \(t=\frac{x + 8}{5}\) into \(y = 2t^{2}+6\). Then \(y=2(\frac{x + 8}{5})^{2}+6\)
Expand \((\frac{x + 8}{5})^{2}=\frac{x^{2}+16x + 64}{25}\)
So \(y = 2\times\frac{x^{2}+16x + 64}{25}+6=\frac{2x^{2}+32x+128}{25}+6\)
Since \(6=\frac{150}{25}\), then \(y=\frac{2x^{2}+32x + 128+150}{25}=\frac{2x^{2}+32x+278}{25}=\frac{2}{25}x^{2}+\frac{32}{25}x+\frac{278}{25}\)

Answer:

\(y=\frac{2}{25}x^{2}+\frac{32}{25}x+\frac{278}{25}\) (the first option)