Elimination

• July 24th 2014, 04:28 AM
rainbowstar
Elimination
Hey Guys,
I am not sure how would you go about doing question 14? Anyone got any ideas?

Attachment 31336
• July 24th 2014, 04:35 AM
ebaines
Re: Elimination
Strangely worded, but I think they want you to express y as a function of x. For example in the first one if you use the identity $\sin \theta = \sqrt {1- \cos^2 \theta}$ you can get an expression for y(x) which has no $\theta$ term in it.
• July 24th 2014, 04:48 AM
rainbowstar
Re: Elimination
That seems like a possible way. However, the answer is:
x square/a square + y square/bsquare = 1
Could you solve these trig equations simultaneously possibly?How would that work?
• July 24th 2014, 05:03 AM
dennydengler
Re: Elimination
Trigonometric identities needed for these 2 questions:
(a) cos^2(theta) + sin^2(theta) = 1. With that in mind, cos(theta) = x/a and sin(theta) = y/b. That's how you have (x/a)^2+(y/b)^2 = 1

(a) tan^2(theta) + 1 = sec^2(theta). With that in mind, tan(theta) = x/a and sec(theta) = y/b. Then you have (x/a)^2 + 1 = (y/b)^2.

Good day.
• July 24th 2014, 05:03 AM
ebaines
Re: Elimination
Did you try the approach I suggested? It gets to the same answer:

$y = b \sin \theta$

$y = b \sqrt {1- \cos^2 \theta}$

$y^2 = b^2(1- \cos^2 \theta)$

$y^2 = b^2(1 - (\frac x a)^2)$

$\frac {y^2}{b^2} + \frac {x^2}{a^2} = 1$
• July 24th 2014, 05:32 AM
rainbowstar
Re: Elimination
Thanks ebaines and dennydengler