How do you convert $3 = - x^{2} + 3 x y - 4 y^{2}$ $3= -x^2+3xy-4y^2$ into a polar equation?

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Answer 1 · 2016-12-21T12:34:53

$r = \frac{1}{3} \sqrt{(cos θ - sin θ) (2 sin θ - cos θ)}$ $r = 1/3sqrt((cos theta-sin theta)(2 sin theta - cos theta))$ .

$- 3 = (x - 2 y) (x - y)$ $-3=(x-2y)(x-y)$ represents a hyperbola having asymptotes

x = 2y and x = y#

Use conversion formula $(x, y) = r (cos θ, sin θ)$ $(x, y)= r (cos theta, sin theta )$ .

The polar form is

$3 = (\frac{- {cos}^{2} θ + 3 sin θ cos θ - 4 {sin}^{2} θ}{r^{2}}$ $3= ((-cos^2theta+3sin theta cos theta -4 sin^2theta)/r^2$

$= - (cos θ - 2 sin θ) (cos θ - sin θ) \frac{)}{r^{2}}$ $=-(cos theta - 2 sin theta)(cos theta - sin theta))/r^2$

Explicitly,

$r = \frac{1}{3} \sqrt{(cos θ - sin θ) (2 sin θ - cos θ)}$ $r = 1/3sqrt((cos theta-sin theta)(2 sin theta - cos theta))$

The asymptotes are now obtained using r = 0 at the ( meet of the

asymptotes ) center..

So, they are ( for pairs of opposite directions ) $θ = \frac{π}{4}, \frac{5}{4} π$ $theta = pi/4, 5/4pi$

and $θ = π + {tan}^{- 1} (\frac{1}{2}), π + {tan}^{- 1} (\frac{1}{2})$ $theta = pi+tan^(-1)(1/2), pi + tan^(-1)(1/2)$

Note that $θ$ $theta$ for the hyperbola $\in ({tan}^{- 1} (\frac{1}{2}), \frac{π}{4})$ $in (tan^(-1)(1/2), pi/4)$ and

$(π + {tan}^{- 1} (\frac{1}{2}), \frac{5}{4} π)$ $(pi+tan^(-1)(1/2), 5/4pi)$ , for the respective branches, in $Q_{1} and Q_{3}$ $Q_1 and Q_3$ .

graph{(x-y)(x-2y)+3=0 [-40, 40, -20, 20]}

How do you convert 3= -x^2+3xy-4y^2 3=−x2+3xy−4y23= -x^2+3xy-4y^2 into a polar equation?