Respuesta :
Answer:
The curve is a circle of radius 9 centered at the point (0,9) and the equation is [tex]x^2+(y-9)^2 = 81[/tex]
Step-by-step explanation:
Proceed as follows:
[tex] r\cos(\theta) = 9 \sin(2\theta)[/tex]
Take [tex]\sin(2\theta) = 2\sin(\theta)\cos(\theta)[/tex]. Then
[tex] r\cos(\theta) = 9 \sin(2\theta)= 18 \sin(\theta) \cos(\theta)[/tex]
Multiply both side by [tex]r^2[/tex]. Then
[tex] r^2\cdot r\cos(\theta) = 18 \cdot r\sin(\theta) \cdot r\cos(\theta)[/tex]
Use the following substitution [tex] x = r\cos(\theta), y = r\sin(\theta), r^2 = x^2+y^2[/tex]. Then
[tex] (x^2+y^2)\cdot (x) = 18 \cdot (x)\cdot(y)[/tex]
By cancelling out x on both sides we get the following equation
[tex]x^2+y^2 = 18y[/tex] or [tex]x^2+y^2-18y =0[/tex]
Recall that given a expression of the form [tex]x^2-bx[/tex] we can complete the square by adding an substracting the amount [tex]\frac{b^2}{4}[/tex]. So, we get [tex] x^2-bx = (x-\frac{b}{2})^2-\frac{b^2}{4}[/tex]. In our case, we will complete the square for y, then
[tex] x^2+y^2-18y = x^2+y^2-18y+(\frac{18}{2})^2-(\frac{18}{2})^2 = 0[/tex]. Then
[tex] x^2+(y-9)^2-81=0[/tex] or
[tex]x^2+(y-9)^2 = 81[/tex].
Recall that the equation of a circle is given by [tex](x-h)^2+(y-k)^2 = r^2[/tex] where (h,k) is the center of the circle and r is the radius. In our case we have h=0, k = 9 and r = 9. So it is a circle of radius 9 centered at the point (0,9)
