Step 1

We have given a cartesian equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). The equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) is a standard form of ellipse

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and generally, we take polar coordinates x as a \(\cos 0\).

Here, a=3. So set \(x=3 \cos 0\). Now, substitute \(x=3 \cos 0\) in \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)

\(\frac{(3\cos 0)^{2}}{9}+\frac{y^{2}}{4}=1\)

\(\frac{9\cos^{2}0}{9}+\frac{y^{2}}{4}=1\)

Step 2

Use the identity \(1-\cos^{2}0=\sin^{2}0\) to simplify it further.

\(\cos^{2}0+\frac{y^{2}}{4}=1\)

\(\frac{y^{2}}{4}=1-\cos^{2}0\)

\(=\sin^{2}0\)

\(y^{2}=4\sin^{2}0\)

Take square root on both sides to find y. Consider only positive sign.

\(\sqrt{y^{2}}=\pm \sqrt{4\sin^{2}0}\)

\(y=\pm 2\sin 0\)

\(y=2\sin 0\) [Take only (+) sign]

Hence, the polar equation is \(x=3 \cos 0, y=2 \sin 0\).

We have given a cartesian equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). The equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) is a standard form of ellipse

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and generally, we take polar coordinates x as a \(\cos 0\).

Here, a=3. So set \(x=3 \cos 0\). Now, substitute \(x=3 \cos 0\) in \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)

\(\frac{(3\cos 0)^{2}}{9}+\frac{y^{2}}{4}=1\)

\(\frac{9\cos^{2}0}{9}+\frac{y^{2}}{4}=1\)

Step 2

Use the identity \(1-\cos^{2}0=\sin^{2}0\) to simplify it further.

\(\cos^{2}0+\frac{y^{2}}{4}=1\)

\(\frac{y^{2}}{4}=1-\cos^{2}0\)

\(=\sin^{2}0\)

\(y^{2}=4\sin^{2}0\)

Take square root on both sides to find y. Consider only positive sign.

\(\sqrt{y^{2}}=\pm \sqrt{4\sin^{2}0}\)

\(y=\pm 2\sin 0\)

\(y=2\sin 0\) [Take only (+) sign]

Hence, the polar equation is \(x=3 \cos 0, y=2 \sin 0\).