Prove that the segment on the Ox axis cut between

Prove that the segment on the $Ox$ axis cut between the tangent and the normal to the curve $x=2a(\ln\sin t-\sin^{2}t),\ y=a\sin2t$ at an arbitrary point $M$ has length $2a$.

Step 1: Derivatives and slope

\[ \dot x = 2a\cos t\!\left(\frac1{\sin t}-2\sin t\right) = \frac{2a\cos t\cos2t}{\sin t}, \qquad \dot y = 2a\cos2t. \tag{1} \]

\[ k_{t} = \frac{\dot y}{\dot x} = \frac{2a\cos2t\cdot\sin t}{2a\cos t\cos2t} = \tan t, \qquad k_{n} = -\cot t. \tag{2} \]

Step 2: $x$-intercept of the tangent

With $x_{0}=2a(\ln\sin t-\sin^{2}t)$, $y_{0}=a\sin2t=2a\sin t\cos t$, setting $y=0$ in the tangent equation and using $y_{0}/k_{t}=2a\cos^{2}t$: \[x_{T} = x_{0} - \frac{y_{0}}{k_{t}} = 2a\ln\sin t - 2a\sin^{2}t - 2a\cos^{2}t = 2a\ln\sin t - 2a. \tag{3}\]

Step 3: $x$-intercept of the normal

Similarly, with $y_{0}/k_{n}=-2a\sin^{2}t$: \[x_{N} = x_{0} - \frac{y_{0}}{k_{n}} = 2a\ln\sin t - 2a\sin^{2}t + 2a\sin^{2}t = 2a\ln\sin t. \tag{4}\]

Step 4: The segment length

\[ \overline{TN} = |x_{N}-x_{T}| = \bigl|2a\ln\sin t - (2a\ln\sin t - 2a)\bigr| = 2a. \tag{5} \]

Step 5: Verification

\[t=1.0,\ a=1\text{: } x_T=2\ln\sin1-2\approx-2.6353,\ x_N=2\ln\sin1\approx-0.6353,\ |x_N-x_T|=2\ \checkmark\]