Find the Cartesian coordinates of the point on the curve

Find the Cartesian coordinates of the point on the curve $\rho=a(1-\cos\theta)$ that divides the arc length above the polar axis in the ratio $2:3$.

Above the polar axis corresponds to $\theta\in[0,\pi]$; at $\theta=0$ the curve has its cusp, at $\theta=\pi$ its farthest point.

Step 1: Arc-length element

\[\rho'=a\sin\theta,\qquad \sqrt{\rho^2+(\rho')^2}=a\sqrt{2-2\cos\theta}=a\sqrt{4\sin^2\tfrac\theta2}=2a\sin\frac\theta2\]

Step 2: Total arc length above the axis

\[L_{\text{total}}=\int_0^\pi2a\sin\frac\theta2\,d\theta = 4a\left[-\cos\frac\theta2\right]_0^\pi\Big/1 = 4a\]

Step 3: Required partial arc length

Ratio $2:3$ means the first piece is $\frac25$ of the total: $L_1=\frac25\cdot4a=\frac{8a}5$.

Step 4: Solve for the dividing angle $\theta_0$

\[4a\left(1-\cos\frac{\theta_0}2\right)=\frac{8a}5 \implies \cos\frac{\theta_0}2=\frac35,\qquad \sin\frac{\theta_0}2=\frac45\]

Step 5: Find $\rho_0$ at $\theta_0$

\[\cos\theta_0=2\cos^2\frac{\theta_0}2-1=2\cdot\frac9{25}-1=-\frac7{25}\]

\[\rho_0=a(1-\cos\theta_0)=a\left(1+\frac7{25}\right)=\frac{32a}{25}\]

Step 6: Convert to Cartesian coordinates

Using $\sin\theta_0=2\sin\frac{\theta_0}2\cos\frac{\theta_0}2=2\cdot\frac45\cdot\frac35=\frac{24}{25}$:

\[x=\rho_0\cos\theta_0=\frac{32a}{25}\cdot\left(-\frac7{25}\right)=-\frac{224a}{625}\]

\[y=\rho_0\sin\theta_0=\frac{32a}{25}\cdot\frac{24}{25}=\frac{768a}{625}\]

\[\boxed{(x,y)=\left(-\frac{224a}{625},\,\frac{768a}{625}\right)}\]