Question

Show that the polar equation for \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(r^{2}=\frac{b^{2}}{1-e^{2} \cos ^{2} \theta} \cdot \quad\) Ellipse

Short answer

After converting the Cartesian to polar coordinates and simplifying, it's confirmed that the polar equation for \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is indeed \(r^{2}=\frac{b^{2}}{1-e^{2} \cos ^{2} \theta}\).

Step-by-step solution

Convert Cartesian to Polar Coordinates

The conversion formulas between Cartesian and polar coordinates are \(x = r \cos \theta\) and \(y = r \sin \theta\). We substitute these into the equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\) to get \(\frac{(r \cos \theta)^2}{a^2} + \frac{(r \sin \theta)^2}{b^2} = 1\).

Simplify the Equation

Rewrite our obtained equation in a simple form by opening brackets to get \( \frac{r^2 \cos ^{2} \theta}{a^2} + \frac{r^2 \sin ^{2} \theta}{b^2} =1\).

Further Simplification

Since \(\sin^2θ + \cos^2θ = 1\), this equation can be written as \(r^2 \left(\frac{\cos ^{2} \theta}{a^2} + \frac{\sin ^{2} \theta}{b^2}\right) =1\). We can solve for \(r^2\) to find \(r^2 = \frac{b^2}{1-e^{2} \cos ^{2} \theta}\) given that \(e^2 = 1 - (b^2/a^2)\).