conics

**sux@math** · May 19th 2008, 04:54 PM

Hey Gang:
I have my final exam tomorrow night and I have "preparation test" I was given, so I have been finding like example questions to quiz myself - here are my weakspots that I can quite get, as there are no samples in all of my textbooks that the same "type":

1. State the sum of focal radii for the ellipse with the equation:

$\frac{x^2}{16}+\frac{y^2}{9}=1$

I get 2 different answers via this:

$a^2=16$ , therefore, a=4
$b^2=9$ , therefore, b=3
$c^2=a^2+b^2$
$c^2=16+9$
$c^2=25$ , therefore c=5, therefore the foci=5, therefore sum of focal radii=10

OR, according to my book, sum of focal radii=2a, therfore 2(4)=8?

2. State the equation of the directrix for the parabola with the vertex (0,0) and focus (0, -3).

$x^2=4py$
$x^2=4(-3)$

3. Solve for x: $3+9^{1-2x}$

I have NO idea how to approach this one. I have never seen a variable *in* an exponent. A hint or two, please on how to start? (I am probably just mental-blocking, so a push in the right direction will most likely make me go "AHH!")

Thanks for all & any help!

**Mathstud28** · May 19th 2008, 05:02 PM

Originally Posted by sux@math

Hey Gang:
I have my final exam tomorrow night and I have "preparation test" I was given, so I have been finding like example questions to quiz myself - here are my weakspots that I can quite get, as there are no samples in all of my textbooks that the same "type":

1. State the sum of focal radii for the ellipse with the equation:

$\frac{x^2}{16}+\frac{y^2}{9}=1$

I get 2 different answers via this:

$a^2=16$ , therefore, a=4
$b^2=9$ , therefore, b=3
$c^2=a^2+b^2$
$c^2=16+9$
$c^2=25$ , therefore c=5, therefore the foci=5, therefore sum of focal radii=10

OR, according to my book, sum of focal radii=2a, therfore 2(4)=8?

2. State the equation of the directrix for the parabola with the vertex (0,0) and focus (0, -3).

$x^2=4py$
$x^2=4(-3)$

3. Solve for x: $3+9^{1-2x}$

I have NO idea how to approach this one. I have never seen a variable *in* an exponent. A hint or two, please on how to start? (I am probably just mental-blocking, so a push in the right direction will most likely make me go "AHH!")

Thanks for all & any help!

I dont like conics.....so

$3+9^{1-2x}=0$ I am assuming the 0

so then $9^{1-2x}=-3$ ...o wait it cant be zero can it haha....ok so what is the real equation

**sux@math** · May 19th 2008, 05:10 PM

Oh, DUH! I can't believe I forgot the whole thing -

#3 is
$3+9^{1-2x}=30$

**Mathstud28** · May 19th 2008, 05:48 PM

Originally Posted by sux@math

Oh, DUH! I can't believe I forgot the whole thing -

#3 is
$3+9^{1-2x}=30$

$3+9^{1-2x}=30\Rightarrow{9^{1-2x}=27=9^{\frac{3}{2}}}$

taking the $\log_9$ of both sides we arrive at

$1-2x=\frac{3}{2}\Rightarrow{\frac{-1}{4}}$

**Isomorphism** · May 20th 2008, 03:34 AM

Originally Posted by sux@math

Hey Gang:
I have my final exam tomorrow night and I have "preparation test" I was given, so I have been finding like example questions to quiz myself - here are my weakspots that I can quite get, as there are no samples in all of my textbooks that the same "type":

1. State the sum of focal radii for the ellipse with the equation:

$\frac{x^2}{16}+\frac{y^2}{9}=1$

I get 2 different answers via this:

$a^2=16$ , therefore, a=4
$b^2=9$ , therefore, b=3
$c^2=a^2+b^2$
$c^2=16+9$
$c^2=25$ , therefore c=5, therefore the foci=5, therefore sum of focal radii=10

OR, according to my book, sum of focal radii=2a, therfore 2(4)=8?

Mostly you have messed up the definition of focal radii... How is it defined to you?
According to this - Mathwords: Ellipse , your book is right... Anyway its a well know property of an ellipse and in many cases this property is used to distinguish an ellipse from other conics.

2. State the equation of the directrix for the parabola with the vertex (0,0) and focus (0, -3).

$x^2=4py$
$x^2=4(-3)$

[IMG]file:///C:/DOCUME%7E1/THEONE%7E1/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/THEONE%7E1/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]First of all since the focus is on the x axis and the vertex is the origin... its a vertical parabola opening downwards. So the equation is of the form
$x^2 =- 4py$ , where p is the distance of the vertex from the focus. Here p = 3. Hence the equation is
$x^2 = -12y$

So you are right(except for a typo)

Thanks for all & any help![/quote]

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