Q6: Which is the graph of the polar equation r=1/1-sin theta.



Answer:
The correct answer is C.
Step-by-step explanation:
The given equation is;
[tex]r=\frac{1}{1-\sin(\theta)}[/tex]
This implies that;
[tex]r(1-\sin(\theta)=1[/tex]
[tex]r-r\sin(\theta)=1[/tex]
Let us write in Cartesian coordinates by substituting;
[tex]r=\sqrt{x^2+y^2} ,y=r\sin(\theta)[/tex]
[tex]\sqrt{x^2+y^2}-y=1[/tex]
[tex]\sqrt{x^2+y^2}=y+1[/tex]
Square both sides;
[tex](\sqrt{x^2+y^2})^2=(y+1)^2[/tex]
This implies that;
[tex]x^2+y^2=y^2+2y+1[/tex]
[tex]x^2=2y+1[/tex]
[tex]y=\frac{1}{2}x^2-\frac{1}{2}[/tex]
This is an equation of a parabola that opens upwards with a y-intercept of [tex]-\frac{1}{2}[/tex].
The correct choice is C