Convert to Polar y=6x^2

$y = 6 x^{2}$

Since $x = r \cos (θ)$ , replace $x$ with $r \cos (θ)$ .

$y = 6 {(r \cos (θ))}^{2}$

Since $y = r \sin (θ)$ , replace $y$ with $r \sin (θ)$ .

$r \sin (θ) = 6 {(r \cos (θ))}^{2}$

Solve for $r$ .

Subtract $6 {(r \cos (θ))}^{2}$ from both sides of the equation.

$r \sin (θ) - 6 {(r \cos (θ))}^{2} = 0$

Apply the product rule to $r \cos (θ)$ .

$r \sin (θ) - 6 r^{2} \cos^{2} (θ) = 0$

Factor $r$ out of $r \sin (θ) - 6 r^{2} \cos^{2} (θ)$ .

Factor $r$ out of $r \sin (θ)$ .

$r (\sin (θ)) - 6 r^{2} \cos^{2} (θ) = 0$

Factor $r$ out of $- 6 r^{2} \cos^{2} (θ)$ .

$r (\sin (θ)) + r (- 6 r \cos^{2} (θ)) = 0$

Factor $r$ out of $r (\sin (θ)) + r (- 6 r \cos^{2} (θ))$ .

$r (\sin (θ) - 6 r \cos^{2} (θ)) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$r = 0$

$\sin (θ) - 6 r \cos^{2} (θ) = 0$

Set $r$ equal to $0$ .

$r = 0$

Set $\sin (θ) - 6 r \cos^{2} (θ)$ equal to $0$ and solve for $r$ .

Set $\sin (θ) - 6 r \cos^{2} (θ)$ equal to $0$ .

$\sin (θ) - 6 r \cos^{2} (θ) = 0$

Solve $\sin (θ) - 6 r \cos^{2} (θ) = 0$ for $r$ .

Subtract $\sin (θ)$ from both sides of the equation.

$- 6 r \cos^{2} (θ) = - \sin (θ)$

Divide each term in $- 6 r \cos^{2} (θ) = - \sin (θ)$ by $- 6 \cos^{2} (θ)$ and simplify.

Divide each term in $- 6 r \cos^{2} (θ) = - \sin (θ)$ by $- 6 \cos^{2} (θ)$ .

$\frac{- 6 r \cos^{2} (θ)}{- 6 \cos^{2} (θ)} = \frac{- \sin (θ)}{- 6 \cos^{2} (θ)}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 r \cos^{2} (θ)}{- 6 \cos^{2} (θ)} = \frac{- \sin (θ)}{- 6 \cos^{2} (θ)}$

Rewrite the expression.

$\frac{r \cos^{2} (θ)}{\cos^{2} (θ)} = \frac{- \sin (θ)}{- 6 \cos^{2} (θ)}$

Cancel the common factor of $\cos^{2} (θ)$ .

Cancel the common factor.

$\frac{r \cos^{2} (θ)}{\cos^{2} (θ)} = \frac{- \sin (θ)}{- 6 \cos^{2} (θ)}$

Divide $r$ by $1$ .

$r = \frac{- \sin (θ)}{- 6 \cos^{2} (θ)}$

Simplify the right side.

Factor $\cos (θ)$ out of $\cos^{2} (θ)$ .

$r = \frac{- \sin (θ)}{- 6 (\cos (θ) \cos (θ))}$

Separate fractions.

$r = \frac{- 1}{- 6 (\cos (θ))} \cdot \frac{\sin (θ)}{\cos (θ)}$

Convert from $\frac{\sin (θ)}{\cos (θ)}$ to $\tan (θ)$ .

$r = \frac{- 1}{- 6 (\cos (θ))} \tan (θ)$

Separate fractions.

$r = \frac{- 1}{- 6} \cdot \frac{1}{\cos (θ)} \tan (θ)$

Convert from $\frac{1}{\cos (θ)}$ to $\sec (θ)$ .

$r = \frac{- 1}{- 6} \sec (θ) \tan (θ)$

Dividing two negative values results in a positive value.

$r = \frac{1}{6} \sec (θ) \tan (θ)$

Combine $\frac{1}{6}$ and $\sec (θ)$ .

$r = \frac{\sec (θ)}{6} \tan (θ)$

Combine $\frac{\sec (θ)}{6}$ and $\tan (θ)$ .

$r = \frac{\sec (θ) \tan (θ)}{6}$

The final solution is all the values that make $r (\sin (θ) - 6 r \cos^{2} (θ)) = 0$ true.

$r = 0$

$r = \frac{\sec (θ) \tan (θ)}{6}$

$r = 0$

$r = \frac{\sec (θ) \tan (θ)}{6}$

Convert to Polar y=6x^2

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