Convert to Polar x=(y^2)/5

$x = \frac{y^{2}}{5}$

Step 1

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

$r \cos (θ) = \frac{y^{2}}{5}$

Step 2

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

$r \cos (θ) = \frac{{(r \sin (θ))}^{2}}{5}$

Step 3

Solve for $r$ .

Multiply both sides by $5$ .

$r \cos (θ) \cdot 5 = \frac{{(r \sin (θ))}^{2}}{5} \cdot 5$

Simplify.

Simplify the left side.

Move $5$ to the left of $r \cos (θ)$ .

$5 r \cos (θ) = \frac{{(r \sin (θ))}^{2}}{5} \cdot 5$

Simplify the right side.

Simplify $\frac{{(r \sin (θ))}^{2}}{5} \cdot 5$ .

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

$5 r \cos (θ) = \frac{r^{2} \sin^{2} (θ)}{5} \cdot 5$

Cancel the common factor of $5$ .

Cancel the common factor.

$5 r \cos (θ) = \frac{r^{2} \sin^{2} (θ)}{5} \cdot 5$

Rewrite the expression.

$5 r \cos (θ) = r^{2} \sin^{2} (θ)$

Solve for $r$ .

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

$5 r \cos (θ) - r^{2} \sin^{2} (θ) = 0$

Factor $r$ out of $5 r \cos (θ) - r^{2} \sin^{2} (θ)$ .

Factor $r$ out of $5 r \cos (θ)$ .

$r (5 \cos (θ)) - r^{2} \sin^{2} (θ) = 0$

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

$r (5 \cos (θ)) + r (- r \sin^{2} (θ)) = 0$

Factor $r$ out of $r (5 \cos (θ)) + r (- r \sin^{2} (θ))$ .

$r (5 \cos (θ) - r \sin^{2} (θ)) = 0$

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

$r = 0$

$5 \cos (θ) - r \sin^{2} (θ) = 0$

Set $r$ equal to $0$ .

$r = 0$

Set $5 \cos (θ) - r \sin^{2} (θ)$ equal to $0$ and solve for $r$ .

Set $5 \cos (θ) - r \sin^{2} (θ)$ equal to $0$ .

$5 \cos (θ) - r \sin^{2} (θ) = 0$

Solve $5 \cos (θ) - r \sin^{2} (θ) = 0$ for $r$ .

Subtract $5 \cos (θ)$ from both sides of the equation.

$- r \sin^{2} (θ) = - 5 \cos (θ)$

Divide each term in $- r \sin^{2} (θ) = - 5 \cos (θ)$ by $- 1 \sin^{2} (θ)$ and simplify.

Divide each term in $- r \sin^{2} (θ) = - 5 \cos (θ)$ by $- 1 \sin^{2} (θ)$ .

$\frac{- r \sin^{2} (θ)}{- 1 \sin^{2} (θ)} = \frac{- 5 \cos (θ)}{- 1 \sin^{2} (θ)}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{r \sin^{2} (θ)}{\sin^{2} (θ)} = \frac{- 5 \cos (θ)}{- 1 \sin^{2} (θ)}$

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

Cancel the common factor.

$\frac{r \sin^{2} (θ)}{\sin^{2} (θ)} = \frac{- 5 \cos (θ)}{- 1 \sin^{2} (θ)}$

Divide $r$ by $1$ .

$r = \frac{- 5 \cos (θ)}{- 1 \sin^{2} (θ)}$

Simplify the right side.

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

$r = \frac{- 5 \cos (θ)}{- 1 (\sin (θ) \sin (θ))}$

Separate fractions.

$r = \frac{- 5}{- 1 (\sin (θ))} \cdot \frac{\cos (θ)}{\sin (θ)}$

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

$r = \frac{- 5}{- 1 (\sin (θ))} \cot (θ)$

Separate fractions.

$r = \frac{- 5}{- 1} \cdot \frac{1}{\sin (θ)} \cot (θ)$

Convert from $\frac{1}{\sin (θ)}$ to $\csc (θ)$ .

$r = \frac{- 5}{- 1} \csc (θ) \cot (θ)$

Divide $- 5$ by $- 1$ .

$r = 5 \csc (θ) \cot (θ)$

The final solution is all the values that make $r (5 \cos (θ) - r \sin^{2} (θ)) = 0$ true.

$r = 0$

$r = 5 \csc (θ) \cot (θ)$

$r = 0$

$r = 5 \csc (θ) \cot (θ)$

$r = 0$

$r = 5 \csc (θ) \cot (θ)$

Convert to Polar x=(y^2)/5

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