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
No example? No formula? No theory? You've been taught nothing?
Read this: Rotation of Axes - How Conic Sections Transform Under Rotations
Multiply both sides of the equation by $\displaystyle 6+9\cos(\vartheta)$
We now have $\displaystyle 6r+9r\cos(\vartheta)=18$
Now, we convert to rectangular: $\displaystyle 6\sqrt{x^2+y^2}+9x=18$
Thus, $\displaystyle 6\sqrt{x^2+y^2}=18-9x$
Square both sides:
$\displaystyle 4(x^2+y^2)=9(4-4x+x^2)$
Now we do some manipulating
$\displaystyle \bigg(x-\frac{36}{5}\bigg)^2-\frac{4}{5}y^2=-\frac{180}{25}+\frac{64}{25}$
$\displaystyle \bigg(x-\frac{36}{5}\bigg)^2-\frac{4}{5}y^2=-\frac{116}{25}$
$\displaystyle \therefore \color{red}\boxed{\frac{y^2}{\left(\frac{29}{5}\ri ght)}-\frac{\left(x-\frac{36}{5}\right)^2}{\frac{116}{5}}=1}$
If I'm not mistaken, this is a hyperbola...
--Chris
Please post new questions on new threads.
You're killing yourselves, here, folks.
$\displaystyle r = \frac{18}{6 + 9 \cos(\theta)}$
$\displaystyle r = \frac{3}{1 + \frac{3}{2} \cos(\theta)}$
$\displaystyle r = \frac{\frac{3}{2} (2)}{1 + \frac{3}{2} \cos(\theta)}$
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?