Respuesta :
For converting polar to cartesian form we know
[tex]r = \sqrt{x^2 + y^2} [/tex]
[tex]x = rcos \theta , y = rsin \theta[/tex]
Given equation is
[tex]r = \frac{-72}{12} + 6sin \theta [/tex]
We can simplify it as
[tex]r = -6 + 6sin \theta[/tex]
Now we can write it as
[tex]r^2 = -6r + 6rsin \theta[/tex]
Now we can use
[tex]r^2 = x^ 2+ y^2 , rsin \theta = y , r = \sqrt{x^2 + y^2} [/tex]
So equation we can write it as
[tex]x^2 + y^2 = -6 \sqrt{x^2 + y^2} + 6y [/tex]
[tex]x^2 + y^2 - 6y = -6 \sqrt{x^2 + y^2} [/tex]
On squaring both sides
[tex](x^2 + y^2 - 6y)^2 = (-6 \sqrt{x^2 + y^2})^2 [/tex]
[tex]x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3 -12x^2y = 36(x^2 + y^2)[/tex]
[tex]x^4 + y^4 +36y^2 + 2x^2y^2 -12y^3 -12x^2y = 36x^2 + 36y^2[/tex]
[tex]x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3-12x^2y - 36x^2 - 36y^2 = 0[/tex]
[tex]x^4 + y^4 -12y^3 + 2x^2y^2 - 12x^2y - 36x^2 = 0[/tex]
[tex]r = \sqrt{x^2 + y^2} [/tex]
[tex]x = rcos \theta , y = rsin \theta[/tex]
Given equation is
[tex]r = \frac{-72}{12} + 6sin \theta [/tex]
We can simplify it as
[tex]r = -6 + 6sin \theta[/tex]
Now we can write it as
[tex]r^2 = -6r + 6rsin \theta[/tex]
Now we can use
[tex]r^2 = x^ 2+ y^2 , rsin \theta = y , r = \sqrt{x^2 + y^2} [/tex]
So equation we can write it as
[tex]x^2 + y^2 = -6 \sqrt{x^2 + y^2} + 6y [/tex]
[tex]x^2 + y^2 - 6y = -6 \sqrt{x^2 + y^2} [/tex]
On squaring both sides
[tex](x^2 + y^2 - 6y)^2 = (-6 \sqrt{x^2 + y^2})^2 [/tex]
[tex]x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3 -12x^2y = 36(x^2 + y^2)[/tex]
[tex]x^4 + y^4 +36y^2 + 2x^2y^2 -12y^3 -12x^2y = 36x^2 + 36y^2[/tex]
[tex]x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3-12x^2y - 36x^2 - 36y^2 = 0[/tex]
[tex]x^4 + y^4 -12y^3 + 2x^2y^2 - 12x^2y - 36x^2 = 0[/tex]
Answer:
The answer is b. [tex]4x^2+3y^2-24y-144=0[/tex]
Step-by-step explanation:
The equation is:
[tex]r=\frac{-72}{12+6\sin(\theta)}[/tex]
Cross multiply;:
[tex]r(12+6\sin(\theta)=-72[/tex]
Expand;:
[tex]12r+6r\sin(\theta)=-72[/tex]
Divide through by 6;:
[tex]2r+r\sin(\theta)=-12[/tex]
[tex]2r=-12-r\sin(\theta)[/tex]
Substitute;:
[tex]y=r\sin(\theta), r=\sqrt{x^2+y^2}[/tex]
[tex]2\sqrt{x^2+y^2}=-12-y[/tex]
Square both sides;:
[tex](2\sqrt{x^2+y^2})^2=(-12-y)^2[/tex]
[tex]4(x^2+y^2)=144+24y+y^2[/tex]
[tex]4x^2+4y^2=144+24y+y^2[/tex]
Simplify;:
[tex]4x^2+3y^2-24y-144=0[/tex]
