Show that 4 rho²=3p²+4a² for the curve x=dfrac a2(3 cos

Show that $4\rho^2=3p^2+4a^2$ for the curve $x=\dfrac a2(3\cos t+\cos3t),\ y=\dfrac a2(3\sin t+\sin3t)$, where $\rho$ is the distance from the origin to the point and $p$ is the distance from the origin to the tangent line there.

Step 1: Velocities

\[ \dot x = -3a\sin2t\cos t, \qquad \dot y = 3a\cos2t\cos t, \tag{1} \] using $\sin t+\sin3t=2\sin2t\cos t$ and $\cos t+\cos3t=2\cos2t\cos t$.

Step 2: Compute $\rho^2=x^2+y^2$

\[ \rho^2 = \frac{a^2}4\Bigl[(3\cos t+\cos3t)^2+(3\sin t+\sin3t)^2\Bigr] = \frac{a^2}4\bigl[10+6\cos2t\bigr] = \frac{a^2}2(5+3\cos2t), \tag{2} \] using $\cos t\cos3t+\sin t\sin3t=\cos2t$.

Step 3: The distance from the origin to the tangent

For a parametric curve, $p=\dfrac{|x\dot y-y\dot x|}{\sqrt{\dot x^2+\dot y^2}}$.

Step 4: The numerator $x\dot y-y\dot x$

\[ x\dot y-y\dot x = \frac{3a^2\cos t}2\Bigl[(3\cos t+\cos3t)\cos2t+(3\sin t+\sin3t)\sin2t\Bigr]. \tag{3} \]

The bracket equals $3\cos(2t-t)+\cos(3t-2t)=3\cos t+\cos t=4\cos t$, so

\[ x\dot y-y\dot x = \frac{3a^2\cos t}{2}\cdot4\cos t = 6a^2\cos^2t. \tag{4} \]

Step 5: The denominator

\[ \dot x^2+\dot y^2 = 9a^2\cos^2t(\sin^22t+\cos^22t) = 9a^2\cos^2t \;\Longrightarrow\; \sqrt{\dot x^2+\dot y^2}=3a|\cos t|. \tag{5} \]

Step 6: Distance $p$

\[ p = \frac{6a^2\cos^2t}{3a|\cos t|} = 2a|\cos t| \;\Longrightarrow\; p^2 = 4a^2\cos^2t. \tag{6} \]

Step 7: Verify the identity

Left side, from (2): $4\rho^2=2a^2(5+3\cos2t)=10a^2+6a^2\cos2t$. Right side, from (6) and $\cos2t=2\cos^2t-1$: \[3p^2+4a^2 = 12a^2\cos^2t+4a^2 = 6a^2(1+\cos2t)+4a^2 = 10a^2+6a^2\cos2t. \tag{7}\]

The two sides agree identically. $\blacksquare$

Step 8: Verification

\[a=2,\ t=0.9\text{: } \rho^2\approx14.72799,\ p^2\approx4.85280,\ 4\rho^2\approx58.91196,\ 3p^2+4a^2\approx58.91196\ \checkmark\]