To convert the parametric equations [tex]\( y = \frac{t + 9}{t} \)[/tex] and [tex]\( x = \frac{4}{\sqrt{t}} \)[/tex] into rectangular form, we need to eliminate the parameter [tex]\( t \)[/tex].
Let's start with the equation for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{4}{\sqrt{t}} \][/tex]
First, solve for [tex]\( t \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ x \sqrt{t} = 4 \][/tex]
Square both sides to get rid of the square root:
[tex]\[ x^2 t = 16 \][/tex]
Now, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{16}{x^2} \][/tex]
Next, substitute this expression for [tex]\( t \)[/tex] into the equation for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{t + 9}{t} \][/tex]
Substitute [tex]\( t = \frac{16}{x^2} \)[/tex] into the equation:
[tex]\[ y = \frac{\frac{16}{x^2} + 9}{\frac{16}{x^2}} \][/tex]
Combine the fractions in the numerator:
[tex]\[ y = \frac{\frac{16 + 9 x^2}{x^2}}{\frac{16}{x^2}} \][/tex]
Simplify:
[tex]\[ y = \frac{16 + 9x^2}{16} \][/tex]
[tex]\[ y = 1 + \frac{9x^2}{16} \][/tex]
Thus, the rectangular form of the given parametric equations is:
[tex]\[ y = 1 + \frac{9x^2}{16} \][/tex]
### Graphing the Equation
To graph the equation [tex]\( y = 1 + \frac{9x^2}{16} \)[/tex]:
1. It is a parabola that opens upwards since the coefficient of [tex]\( x^2 \)[/tex] is positive.
2. The vertex of the parabola is at (0,1) because the equation is in the form [tex]\( y = 1 + \frac{9}{16}x^2 \)[/tex].
### Domain Restrictions
Consider the original expressions for [tex]\( y \)[/tex] and [tex]\( x \)[/tex] in terms of [tex]\( t \)[/tex]:
[tex]\[ y = \frac{t + 9}{t} \][/tex]
[tex]\[ x = \frac{4}{\sqrt{t}} \][/tex]
To ensure that [tex]\( x \)[/tex] is defined:
1. [tex]\( \sqrt{t} \)[/tex] must be real and positive.
2. Therefore, [tex]\( t > 0 \)[/tex].
Additionally, [tex]\( x \neq 0 \)[/tex] since [tex]\( t = \frac{16}{x^2} \)[/tex] and division by zero is undefined.
The domain restriction can be stated as:
[tex]\[ x \neq 0 \][/tex]
Thus, combining these observations:
- The domain of [tex]\( x \)[/tex] is [tex]\( x \in \mathbb{R} \setminus \{0\} \)[/tex]. In other words, [tex]\( x \)[/tex] can take any real value except zero.
### Final Answer
The rectangular form of the given parametric equations is:
[tex]\[ y = 1 + \frac{9x^2}{16} \][/tex]
The equation’s graph is a parabola opening upwards with its vertex at (0,1).
The domain restriction on [tex]\( x \)[/tex] is:
[tex]\[ x \neq 0 \][/tex]
