$\displaystyle r = \sin ^{2} \theta$

$\displaystyle r^{2} \cdot r = r^{2} \sin ^{2} \theta$

$\displaystyle r^{3} =\left ( r \sin \theta \right ) ^{2}$

$\displaystyle \left ( r^{2} \right )^{\frac{3}{2}} =\left ( r \sin \theta \right ) ^{2}$

$\displaystyle \left (x^{2} +y^{2} \right )^{\frac{3}{2}} =y ^{2}$

$\displaystyle \left [ \left (x^{2} +y^{2} \right )^{\frac{3}{2}} \right ] ^{2} =\left ( y ^{2} \right ) ^{2}$

$\displaystyle \left (x^{2} +y^{2} \right )^{3} = y ^{4}$

or $\displaystyle y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

$\displaystyle x^{2} = 3xy + 1$

$\displaystyle \left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$\displaystyle r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1$

$\displaystyle r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1$

$\displaystyle r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1$

$\displaystyle r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }$

$\displaystyle r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

$\displaystyle r ^{2} + 1 = \cos \theta$

$\displaystyle r\left ( r ^{2} + 1 \right )= r \cos \theta$

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

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

This line has slope $\displaystyle \frac{4-0}{0- (-6)} = \frac{2}{3}$ and $\displaystyle y$-intercept $\displaystyle (0.4)$, so its Cartesian equation is $\displaystyle y = \frac{2}{3}x + 4$.

By substituting, we can rewrite this:

$\displaystyle r \sin \theta = \frac{2}{3}r \cos \theta + 4$

$\displaystyle r \sin \theta - \frac{2}{3}r \cos \theta = 4$

$\displaystyle 3 r \sin \theta - 2 r \cos \theta = 12$

$\displaystyle r \left ( 3 \sin \theta - 2 \cos \theta \right )= 12$

$\displaystyle r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

$\displaystyle x = r \cos \theta = -4 \cos \frac{\pi}{3} = -4 \cdot \frac{1}{2} = -2$

$\displaystyle y = r \sin \theta = -4 \sin \frac{\pi}{3} = -4 \cdot \frac{ \sqrt{3}}{2} = -2 \sqrt{3}$

The point will have rectangular coordinates $\displaystyle \left (- 2, -2 \sqrt{3} \right )$.

We know $\displaystyle \small y=rsin\theta$ and $\displaystyle \small x=rcos\theta$.

Subbing that in to the equation $\displaystyle \small y=x^{2}$ will give us $\displaystyle \small rsin\theta=r^{2}cos^{2}\theta$.

Multiplying both sides by $\displaystyle \small 1/(rcos^{2}\theta)$ gives us 

$\displaystyle \small \small \small \small \small r=sin\theta /cos^{2}\theta=tan\theta/cos\theta=tan\theta\cdot sec\theta$.

To go from polar form to cartesion coordinates, use the following two relations.

$\displaystyle x=rcos(\theta)$

$\displaystyle y=rsin(\theta)$

In this case, our $\displaystyle r$ is $\displaystyle 3$ and our $\displaystyle \theta$ is $\displaystyle \frac{\pi}{3}$.

Plugging those into our relations we get 

$\displaystyle x=\frac{3}{2}$

$\displaystyle y=\frac{3\sqrt3}{2}$, 

which gives us our $\displaystyle (x,y)$ coordinate.

Although not explicitly stated, the problem is asking for the polar coordinates of the point $\displaystyle (3,4)$. To calculate the magnitude, $\displaystyle r$, calculate the following:

$\displaystyle r=\sqrt{x^2+y^2}$

$\displaystyle r=5$

To calculate $\displaystyle \theta$, do the following:

$\displaystyle \theta=tan^{-1}\left(\frac{y}{x}\right)$ in radians. (The problem asks for radians)

$\displaystyle \theta = 0.927$

To calculate the polar coordinate, use

$\displaystyle r=\sqrt{x^2+y^2}$

$\displaystyle r=\sqrt{29}$

$\displaystyle \theta=tan^{-1}\left(\frac{y}{x}\right)=68$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\displaystyle \theta=248$.

We can convert from rectangular form to polar form by using the following identities: $\displaystyle y=r\sin \theta$ and $\displaystyle x=r\cos \theta$. Given $\displaystyle y=2x^{2}$, then $\displaystyle r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$.

$\displaystyle r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$. Dividing both sides by $\displaystyle r\cos \theta$,

$\displaystyle \tan \theta=2r\cos \theta$

$\displaystyle \frac{\tan \theta}{\cos \theta}=2r$

$\displaystyle \tan \theta}{\sec \theta=2r$

$\displaystyle \frac{1}{2}\tan \theta}{\sec \theta=r$