Hello. Thanks for any help in advance.
just having trouble with the method of eliminating θ from the following pairs of equations:
a. x = 2 cosθ, y = 2 sin θ
b. x = b tan, y = a sec θ
The key is to use the many forms of the Pythagorean Identity - in this case $\displaystyle \sin^2{\theta} + \cos^2{\theta} = 1$ and $\displaystyle \tan^2{\theta} + 1 = \sec^2{\theta}$.
(a) $\displaystyle x = 2\cos{\theta}, y = 2\sin{\theta}$
$\displaystyle x^2 = 4\cos^2{\theta}, y^2 = 4\sin^2{\theta}$
$\displaystyle x^2 + y^2 = 4\cos^2{\theta} + 4\sin^2{\theta}$
$\displaystyle x^2 + y^2 = 4(\cos^2{\theta} + \sin^2{\theta})$
$\displaystyle x^2 + y^2 = 4\cdot 1$
$\displaystyle x^2 + y^2 = 4$.
Can you follow a similar process for (b)?
You're on the right track. Remember that $\displaystyle \tan^2{\theta} + 1 = \sec^2{\theta}$, so $\displaystyle \sec^2{\theta} - \tan^2{\theta} = 1$.
If $\displaystyle x = b\tan{\theta}, y = a\sec{\theta}$
then $\displaystyle x^2 = b^2\tan^2{\theta}, y^2 = a^2\sec^2{\theta}$
$\displaystyle a^2x^2 = a^2b^2\tan^2{\theta}, b^2y^2 = a^2b^2\sec^2{\theta}$.
$\displaystyle a^2b^2\sec^2{\theta} - a^2b^2\tan^2{\theta} = b^2y^2 - a^2x^2$
$\displaystyle a^2b^2(\sec^2{\theta} - \tan^2{\theta}) = b^2y^2 - a^2x^2$
$\displaystyle a^2b^2\cdot 1 = b^2y^2 - a^2x^2$
$\displaystyle a^2b^2 = b^2y^2 - a^2x^2$
$\displaystyle 1 = \frac{b^2y^2 - a^2x^2}{a^2b^2}$
$\displaystyle \frac{b^2y^2}{a^2b^2} - \frac{a^2x^2}{a^2b^2} = 1$
$\displaystyle \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.