Elimination technique questions that frankly, I don't understand.

**katrina782** · November 26th 2009, 09:37 PM

1) Eliminate t to give an equation that relates x and y.
x=(e^(2t))+9 and y=e^(7t)

2) Eliminate t to give an equation that relates x and y.
x=tan(t) and y=sec^2(t)-3

3) Show that this equation represents a circle by rearranging it into the centre-radius form of the equation of a circle. State the coordinates of the center and the radius of the circle:

21y^2+19x^2+83=4x-2x^2-84y

LOLWUT?! Help would be soo much appreciated.

**earboth** · November 26th 2009, 10:24 PM

Originally Posted by katrina782

1) Eliminate t to give an equation that relates x and y.
x=(e^(2t))+9 and y=e^(7t)

1. Solve the 2nd equation for t:

$y=e^{7t}~\implies~7t=\ln(y)~\implies~\boxed{t=\fra c17 \ln(y)}$
and plug in this term into the first equation:

$x=\left(e^{2 \cdot \frac17 \ln(y)}\right)+9~\implies~x=\sqrt[7]{y^2}+9~\implies~y^2=(x-9)^7\ ,\ x>9$

2) Eliminate t to give an equation that relates x and y.
x=tan(t) and y=sec^2(t)-3

This one is for you!

3) Show that this equation represents a circle by rearranging it into the centre-radius form of the equation of a circle. State the coordinates of the center and the radius of the circle:

21y^2+19x^2+83=4x-2x^2-84y

...

1. Re-arrange the equation:

$21x^2 -4x+21y^2+84y = -83$

2. Complete the squares:

$21\left(x^2 - \frac4{21} x + \frac{4}{441}\right) +21\left(y^2+4y+4\right) = -83+21 \cdot \frac4{441} + 21 \cdot 4 = \frac{25}{21}$

3. Divide the equation by the leading factor of the brackets on the LHS:

$\left(x-\frac2{21}\right)^2+(y+2)^2=\left(\frac5{21}\right )^2$

4. Determine the coordinates of the centre and the length of the radius.

**katrina782** · November 26th 2009, 10:27 PM

You're dope. Thanks

**HallsofIvy** · November 27th 2009, 02:49 AM

So "you're dope" is different from "you're a dope"?

I would have done (1) slightly differently, solving only for $e^t$ rather than t itself. From the first equation, [/tex]x= e^{2t}+ 9[/tex], $x- 9= e^{2t}= (e^t)^2$ , $e^t= (x-9)^{1/2}$ . From the second equation, $y= e^{7t}= (e^t)^7$ , $e^t= y^{1/7}$ . Putting those together, $(x-9)^{1/2}= y^{1/7}$ , which, taking both sides to the 14th power, is the same as $(x-9)^7= y^2$

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