# Math Help - 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 $t$ in terms of $x$. Then substitute that result into the second equation.

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

3. Originally Posted by Redding1234
For eliminating the parameter:

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

Alternately, get $t$ in terms of $y$. 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 $tan(\theta)= x$ or $\theta= arctan x$. $sec(arctan(x))= sec(\theta)$ 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 $sin^2(\theta)+ cos^2(\theta)= 1$. Divide both sides by $cos^2(\theta)$ to get $\frac{sin^2(\theta)}{cos^2(\theta)}+ 1= \frac{1}{cos^2(\theta)}$ which is the same as $tan^2(\theta)+ 1= sec^2(\theta)$.

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