How do you convert $1 = {(- 2 x - 3)}^{2} + {(y + 1)}^{2}$ $1=(-2x-3)^2+(y+1)^2$ into polar form?

Question

How do you convert $1 = {(- 2 x - 3)}^{2} + {(y + 1)}^{2}$ $1=(-2x-3)^2+(y+1)^2$ into polar form?

1 Answer

Impact of this question

1660 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-10T17:27:17

$4 {(r cos θ + \frac{3}{2})}^{2} + {(r sin θ + 1)}^{2} = 1$ $4(r cos theta + 3/2)^2+(rsintheta+1)^2=1$ .

Explanation:

graph{(2x+3)^2+(y+1)^2-1=0 [-3, 3, -3, 0]}

The equation has the standard form

$\frac{{(x + \frac{3}{2})}^{2}}{{(\frac{1}{2})}^{2}} + \frac{{(Y + 1)}^{2}}{1^{2}} = 1$ $(x+3/2)^2/(1/2)^2+(Y+1)^2/1^2=1$ that represents the ellipse, with

center $C (- \frac{3}{2}, - 1)$ $C(-3/2, -1)$ , major axis along $x = - \frac{3}{2}$ $x =-3/2$ , minor axis

along $y = - 1$ $y = -1$ , semi major axis a = 1,, semi minor axis $b = \frac{1}{2}$ $b = 1/2$ and eccentricity $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{\sqrt{3}}{2} .$ $e= sqrt(1-b^2/a^2)=sqrt3/2.$

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

So, the polar equation is

$4 {(r cos θ + \frac{3}{2})}^{2} + {(r sin θ + 1)}^{2} = 1$ $4(r cos theta + 3/2)^2+(rsintheta+1)^2=1$ .

Science

Math

Humanities

... and beyond

How do you convert $1 = {(- 2 x - 3)}^{2} + {(y + 1)}^{2}$ $1=(-2x-3)^2+(y+1)^2$ into polar form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you convert 1=(-2x-3)^2+(y+1)^21=(−2x−3)2+(y+1)21=(-2x-3)^2+(y+1)^2 into polar form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you convert $1 = {(- 2 x - 3)}^{2} + {(y + 1)}^{2}$ $1=(-2x-3)^2+(y+1)^2$ into polar form?