Question

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

Answer 1

Answer:

The correct answer is C.

Step-by-step explanation:

The given equation is;

$r=\frac{1}{1-\sin(\theta)}$

This implies that;

$r(1-\sin(\theta)=1$

$r-r\sin(\theta)=1$

Let us write in Cartesian coordinates by substituting;

$r=\sqrt{x^2+y^2} ,y=r\sin(\theta)$

$\sqrt{x^2+y^2}-y=1$

$\sqrt{x^2+y^2}=y+1$

Square both sides;

$(\sqrt{x^2+y^2})^2=(y+1)^2$

This implies that;

$x^2+y^2=y^2+2y+1$

$x^2=2y+1$

$y=\frac{1}{2}x^2-\frac{1}{2}$

This is an equation of a parabola that opens upwards with a  y-intercept of $-\frac{1}{2}$.

The correct choice is C