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

Question

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

2 Answers

Impact of this question

2175 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-11T04:39:19

We know the relations

$x = r cos θ and y = r sin θ$ $x=rcostheta and y =rsintheta$

Again $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

where r and $θ$ $theta$ are the polar coordinate of a point having rectangular coordinate $(x, y)$ $(x,y)$

The given equation in rectanglar form is

$y = - 2 x^{2} + 3 x y - 4$ $y=-2x^2+3xy-4$

$\Rightarrow r sin θ = - 2 r^{2} {cos}^{2} θ + 3 r^{2} sin θ cos θ - 4$ $=>rsintheta=-2r^2cos^2theta+3r^2sinthetacostheta-4$

$\Rightarrow r sin θ + 2 r^{2} {cos}^{2} θ - 3 r^{2} sin θ cos θ + 4 = 0$ $=>rsintheta+2r^2cos^2theta-3r^2sinthetacostheta+4=0$

This is the polar form of the given equation.

Answer 2 · 2016-10-11T04:59:33

Substitute $r cos (θ)$ $rcos(theta)$ for $x$ $x$ and $r sin (θ)$ $rsin(theta)$ for $y$ $y$ :
$r = \frac{- sin (θ) \pm \sqrt{{sin}^{2} (θ) + 16 (3 cos (θ) sin (θ) - 2 {cos}^{2} (θ))}}{(6 cos (θ) sin (θ) - 4 {cos}^{2} (θ))}$ $r = (-sin(theta) +-sqrt(sin^2(theta) + 16(3cos(theta)sin(theta) - 2cos^2(theta))))/((6cos(theta)sin(theta) - 4cos^2(theta)))$

Explanation:

Substitute $r cos (θ)$ $rcos(theta)$ for $x$ $x$ and $r sin (θ)$ $rsin(theta)$ for $y$ $y$ :

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

$r^{2} (3 cos (θ) sin (θ) - 2 {cos}^{2} (θ)) - r sin (θ) - 4$ $r^2(3cos(theta)sin(theta) - 2cos^2(theta)) - rsin(theta) - 4$

This a quadratic of the form $a r^{2} + b r + c$ $ar^2 + br + c$ where $a = (3 cos (θ) sin (θ) - 2 {cos}^{2} (θ))$ $a = (3cos(theta)sin(theta) - 2cos^2(theta))$ , $b = - sin (θ)$ $b = -sin(theta)$ , and $c = - 4$ $c = -4$

Using the quadratic formula, $r = \frac{- b \pm \sqrt{b^{2} - 4 (a) (c)}}{2 a}$ $r = (-b +-sqrt(b^2 - 4(a)(c)))/(2a)$ :

$r = \frac{- sin (θ) \pm \sqrt{{sin}^{2} (θ) + 16 (3 cos (θ) sin (θ) - 2 {cos}^{2} (θ))}}{(6 cos (θ) sin (θ) - 4 {cos}^{2} (θ))}$ $r = (-sin(theta) +-sqrt(sin^2(theta) + 16(3cos(theta)sin(theta) - 2cos^2(theta))))/((6cos(theta)sin(theta) - 4cos^2(theta)))$

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Related questions

Impact of this question

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