How do you convert $3 x^{2} + 6 x y - y^{2} = 9$ $3x^2+6xy-y^2=9$ into polar form?

Question

1761 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-16T07:53:24

$r = \frac{3}{\sqrt{3 \cdot {cos}^{2} θ + 6 \cdot sin θ \cdot cos θ - {sin}^{2} θ}}$ $r=3/sqrt(3*cos^2 theta+6*sin theta*cos theta-sin^2 theta)$

Use $x = r cos θ$ $x=r cos theta$ and $y = r sin θ$ $y=r sin theta$

from the given

$3 x^{2} + 6 x y - y^{2} = 9$ $3x^2+6xy-y^2=9$

$3 {(r cos θ)}^{2} + 6 (r cos θ) (r sin θ) - {(r sin θ)}^{2} = 9$ $3(r cos theta)^2+6(r cos theta)(r sin theta)-(r sin theta)^2=9$

$3 r^{2} {cos}^{2} θ + 6 r^{2} \cdot sin θ \cdot cos θ - r^{2} {sin}^{2} θ = 9$ $3r^2 cos^2 theta+6r^2*sin theta*cos theta-r^2 sin^2 theta=9$

$r^{2} (3 {cos}^{2} θ + 6 \cdot sin θ \cdot cos θ - {sin}^{2} θ) = 9$ $r^2(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta)=9$

$r^{2} = \frac{9}{3 {cos}^{2} θ + 6 \cdot sin θ \cdot cos θ - {sin}^{2} θ}$ $r^2=9/(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta)$

$r = \sqrt{\frac{9}{3 {cos}^{2} θ + 6 \cdot sin θ \cdot cos θ - {sin}^{2} θ}}$ $r=sqrt(9/(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta))$

$r = \frac{3}{\sqrt{3 {cos}^{2} θ + 6 \cdot sin θ \cdot cos θ - {sin}^{2} θ}}$ $r=3/sqrt(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta)$

have a nice day... from the Philippines

How do you convert 3x^2+6xy-y^2=93x2+6xy−y2=93x^2+6xy-y^2=9 into polar form?