# Math Help - help with conic question

1. ## help with conic question

Rotate the axes to eliminate the xy-term. Then write the standard form of the equation of the conic relative to x'y'-plane.

3x^2-4xy+1=0

2. Originally Posted by cityismine
Rotate the axes to eliminate the xy-term. Then write the standard form of the equation of the conic relative to x'y'-plane.

3x^2-4xy+1=0
Are you using eigenvalues and eigenvectors, or the the transformation formula? Either way, where are you stuck?

3. This is from the conics section of a pre-calculus textbook, but the book doesn't give too many examples on solving problems like this. So I'm stuck on this one.

4. Originally Posted by cityismine
This is from the conics section of a pre-calculus textbook, but the book doesn't give too many examples on solving problems like this. So I'm stuck on this one.
No example? No formula? No theory? You've been taught nothing?

Read this: Rotation of Axes - How Conic Sections Transform Under Rotations

5. y^2-4x^2=1

6. Originally Posted by cityismine
y^2-4x^2=1

Bravo!

And I bet it felt good doing it all yourself.

7. How do i turn this conic into rectangular form:

r=18/(6+9cos(theta))

8. Originally Posted by cityismine
How do i turn this conic into rectangular form:

r=18/(6+9cos(theta))
Multiply both sides of the equation by $6+9\cos(\vartheta)$

We now have $6r+9r\cos(\vartheta)=18$

Now, we convert to rectangular: $6\sqrt{x^2+y^2}+9x=18$

Thus, $6\sqrt{x^2+y^2}=18-9x$

Square both sides:

$4(x^2+y^2)=9(4-4x+x^2)$

Now we do some manipulating

$\bigg(x-\frac{36}{5}\bigg)^2-\frac{4}{5}y^2=-\frac{180}{25}+\frac{64}{25}$

$\bigg(x-\frac{36}{5}\bigg)^2-\frac{4}{5}y^2=-\frac{116}{25}$

$\therefore \color{red}\boxed{\frac{y^2}{\left(\frac{29}{5}\ri ght)}-\frac{\left(x-\frac{36}{5}\right)^2}{\frac{116}{5}}=1}$

If I'm not mistaken, this is a hyperbola...

--Chris

9. Thanks for showing me the method, I tried following along, but I got a different answer. Am I right?

(x-3.6)^2 / 5.76 - y^2 / 7.2 = 1

You're killing yourselves, here, folks.

$r = \frac{18}{6 + 9 \cos(\theta)}$

$r = \frac{3}{1 + \frac{3}{2} \cos(\theta)}$

$r = \frac{\frac{3}{2} (2)}{1 + \frac{3}{2} \cos(\theta)}$

Eccentricity: e = 3/2 -- This is an hyperbola
Directrix Displacement: p = 2 -- One directrix is at x = 2

Where is the other Directrix?

Where are the Vertices?

What is the length of the Minor Axis?

11. ?? Why won't this LaTeX work?