A class is to eliminate t from the parametric equations x = t^2 + 3 and y = 4t. Beth says that she can write t = sqrt (x-3) to eliminate the parameter. Why is this wrong?
A) She should have added 3 to x, not subtracted
B) She should always solve for t as a function of y
C) She should have taken both the positive and negative square root
D) She should first substitute y for t before solving
looks like B to me
The correct answer is C) She should have taken both the positive and negative square root.
By taking the square root to eliminate the parameter, Beth only considered the positive square root while writing t = sqrt(x-3). However, when solving for t, we need to consider both positive and negative square roots since the original equation contains a quadratic term, t^2.
When solving for t to eliminate the parameter, we should have written t = ±sqrt(x-3) to include both solutions. Therefore, option C is the correct explanation of why Beth's approach is incorrect.
The correct answer is A) She should have added 3 to x, not subtracted.
To understand why Beth's approach is wrong, let's go through the steps she followed:
1) She started with the parametric equations x = t^2 + 3 and y = 4t.
2) Next, she attempted to eliminate the parameter t by solving for t in terms of x: t = sqrt(x-3).
Now, let's see why this method is incorrect.
The original parametric equations are x = t^2 + 3 and y = 4t. In these equations, the variable t represents a parameter, not a variable to solve for.
When we try to eliminate t, we need to express x and y in terms of each other without using t.
Beth's mistake was taking the square root of (x-3) to solve for t. If we substitute this value of t into the equation y = 4t, we get y = 4 * sqrt(x-3). However, this new equation involves both x and y, so it does not eliminate the parameter t.
The correct approach to eliminate t from the parametric equations is to solve one equation for t and substitute it into the other equation.
In this case, we can solve the second equation y = 4t for t: t = y/4.
Then, we substitute this value of t into the first equation x = t^2 + 3:
x = (y/4)^2 + 3.
Simplifying this equation, we get:
x = y^2/16 + 3.
Now, we have expressed x in terms of y without using the parameter t. And this correctly eliminates the parameter t from the parametric equations.