# Thread: Parametric Parameter Elimination

1. ## Parametric Parameter Elimination

Eliminate the parameter to find a cartesian equation of the curve.

x = tan(t)
y = sec^2(t)

from 0 <= t <= pi/2

Thank you!

2. For eliminating the parameter:

Get $\displaystyle t$ in terms of $\displaystyle x$. Then substitute that result into the second equation.

Alternately, get $\displaystyle t$ in terms of $\displaystyle y$. Then substitute that result into the first equation.

3. Originally Posted by Redding1234
For eliminating the parameter:

Get $\displaystyle t$ in terms of $\displaystyle x$. Then substitute that result into the second equation.

Alternately, get $\displaystyle t$ in terms of $\displaystyle y$. Then substitute that result into the first equation.
y= sec^2(arctan(x)), I've already gotten that, but how does one simplify that?

Is there some identity that's slipping my mind right now?

4. You can do it this way: imagine a right triangle with "opposite side" of length x and "near side" of length 1. Then $\displaystyle tan(\theta)= x$ or $\displaystyle \theta= arctan x$. $\displaystyle sec(arctan(x))= sec(\theta)$ and secant is defined as "hypotenuse over near side". You can get the hypotenuse from the Pythagorean theorem.

That's really a "visual" way of getting the following identity:

You know that $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$. Divide both sides by $\displaystyle cos^2(\theta)$ to get $\displaystyle \frac{sin^2(\theta)}{cos^2(\theta)}+ 1= \frac{1}{cos^2(\theta)}$ which is the same as $\displaystyle tan^2(\theta)+ 1= sec^2(\theta)$.

Taking $\displaystyle \theta= arctan(x)$ in that gives gives $\displaystyle sec^2(arctan(x))= tan^2(arctan(x)+ 1= x^2+ 1$. Therefore, $\displaystyle sec(arctan(x))= \sqrt{x^2+ 1}$.

### eliminate the parameter to find a cartesian equation x=tan2theta and y=sec thetha

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