urgent! please help me to solve Find a polar equation for the curve represented by the given Cartesian equation.

Question

Find a polar equation for the curve represented by the given Cartesian equation.

y= − 2x^ 2

Answer 1

y = -2x²

Let x = rcosθ and y = rsinθ

So, rsinθ = -2r²cos²θ

Divide by r to get sinθ = -2rcos²θ.

So, r = (-1/2)sinθ / cos²θ

r = (-1/2)(sinθ/cosθ)(1/cosθ)

So, r = (-1/2)tanθsecθ, where 0 < θ < π/2 or π/2 < θ < π

Answer 2

### Problem:

Find a polar equation for the curve represented by the given Cartesian equation:

\[

y = -2x^2

\]

### Solution:

We will use the polar-coordinate relationships between \(x\), \(y\), and \(r, \theta\):

\[

x = r \cos \theta

\]

\[

y = r \sin \theta

\]

### Step 1: Substituting the expressions for \(x\) and \(y\) in terms of polar coordinates into the Cartesian equation.

The given equation is:

\[

y = -2x^2

\]

Substituting \(x = r \cos \theta\) and \(y = r \sin \theta\), we get:

\[

r \sin \theta = -2 (r \cos \theta)^2

\]

### Step 2: Simplifying the equation.

First, expand the square term:

\[

r \sin \theta = -2 r^2 \cos^2 \theta

\]

### Step 3: Solve for \(r\).

Divide both sides of the equation by \(r\) (assuming \(r \neq 0\)):

\[

\sin \theta = -2 r \cos^2 \theta

\]

Now, solve for \(r\):

\[

r = \frac{\sin \theta}{-2 \cos^2 \theta}

\]

### Final Polar Equation:

\[

r = \frac{\sin \theta}{-2 \cos^2 \theta}

\]

This is the polar equation for the curve represented by the Cartesian equation \(y = -2x^2\).