What the is the polar form of $y = \frac{y^{2}}{x} + (x - 3) (y - 5)$ $y = y^2/x+(x-3)(y-5)$ ?

Question

What the is the polar form of $y = \frac{y^{2}}{x} + (x - 3) (y - 5)$ $y = y^2/x+(x-3)(y-5)$ ?

1 Answer

Impact of this question

1209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-18T12:43:47

$r (- sin θ tan θ - r sin θ cos θ + 4 sin θ + 5 cos θ) = 15$ $r(-sinthetatantheta-rsinthetacostheta+4sintheta+5costheta)=15$

Explanation:

First, we expand everything to get:
$y = \frac{y^{2}}{x} + x y - 3 y - 5 y + 15$ $y=y^2/x+xy-3y-5y+15$

Now we need to use these:
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

$r sin θ = \frac{r^{2} {sin}^{2} θ}{r cos θ} + r cos θ r sin θ - 3 r sin θ - 5 r cos θ + 15$ $rsintheta=(r^2sin^2theta)/(rcostheta)+rcosthetarsintheta-3rsintheta-5rcostheta+15$

$r sin θ = r sin θ tan θ + r^{2} sin θ cos θ - 3 r sin θ - 5 r cos θ + 15$ $rsintheta=rsinthetatantheta+r^2sinthetacostheta-3rsintheta-5rcostheta+15$

$r sin θ - r sin θ tan θ - r^{2} sin θ cos θ + 3 r sin θ + 5 r cos θ = 15$ $rsintheta-rsinthetatantheta-r^2sinthetacostheta+3rsintheta+5rcostheta=15$

$r (- sin θ tan θ - r sin θ cos θ + 4 sin θ + 5 cos θ) = 15$ $r(-sinthetatantheta-rsinthetacostheta+4sintheta+5costheta)=15$

We cannot simplify this any further, so it stays as an implicit polar equation.

Science

Math

Humanities

... and beyond

What the is the polar form of $y = \frac{y^{2}}{x} + (x - 3) (y - 5)$ $y = y^2/x+(x-3)(y-5)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What the is the polar form of y = y^2/x+(x-3)(y-5) y=y2x+(x−3)(y−5)y = y^2/x+(x-3)(y-5) ?

1 Answer

Explanation:

Related questions

Impact of this question

What the is the polar form of $y = \frac{y^{2}}{x} + (x - 3) (y - 5)$ $y = y^2/x+(x-3)(y-5)$ ?