To write the parametric equations [tex]\( y = 2t^2 + 6 \)[/tex] and [tex]\( x = 5t - 8 \)[/tex] in rectangular form, we need to eliminate the parameter [tex]\( t \)[/tex]. Let's do this step by step.
1. Solve for [tex]\( t \)[/tex] in terms of [tex]\( x \)[/tex]:
Given [tex]\( x = 5t - 8 \)[/tex], we solve for [tex]\( t \)[/tex]:
[tex]\[
x + 8 = 5t
\][/tex]
[tex]\[
t = \frac{x + 8}{5}
\][/tex]
2. Substitute [tex]\( t \)[/tex] into the equation for [tex]\( y \)[/tex]:
Substitute [tex]\( t = \frac{x + 8}{5} \)[/tex] into [tex]\( y = 2t^2 + 6 \)[/tex]:
[tex]\[
y = 2 \left( \frac{x + 8}{5} \right)^2 + 6
\][/tex]
3. Simplify the expression:
First, square [tex]\( \frac{x + 8}{5} \)[/tex]:
[tex]\[
\left( \frac{x + 8}{5} \right)^2 = \frac{(x + 8)^2}{25}
\][/tex]
Next, multiply by 2:
[tex]\[
y = 2 \cdot \frac{(x + 8)^2}{25} + 6
\][/tex]
[tex]\[
y = \frac{2(x + 8)^2}{25} + 6
\][/tex]
4. Expand and simplify:
Expand [tex]\( (x + 8)^2 \)[/tex]:
[tex]\[
(x + 8)^2 = x^2 + 16x + 64
\][/tex]
Substitute back into the equation:
[tex]\[
y = \frac{2(x^2 + 16x + 64)}{25} + 6
\][/tex]
Distribute the 2:
[tex]\[
y = \frac{2x^2 + 32x + 128}{25} + 6
\][/tex]
Combine into a single fraction and simplify:
[tex]\[
y = \frac{2x^2 + 32x + 128}{25} + \frac{150}{25}
\][/tex]
[tex]\[
y = \frac{2x^2 + 32x + 128 + 150}{25}
\][/tex]
[tex]\[
y = \frac{2x^2 + 32x + 278}{25}
\][/tex]
Thus, the rectangular form of the given parametric equations is:
[tex]\[
y = \frac{2x^2 + 32x + 278}{25}
\][/tex]
Comparing this result to the given choices:
- [tex]\( y = \frac{2}{25} x^2 + \frac{32}{25} x + \frac{278}{25} \)[/tex]
We see that they match. Therefore, the correct answer is:
[tex]\[
y = \frac{2}{25} x^2 + \frac{32}{25} x + \frac{278}{25}
\][/tex]
