Find the product of the distances from the foci

Find the product of the distances from the foci of the ellipse $x=a\cos t,\ y=b\sin t$ ($a>b>0$) to the normal at an arbitrary point $M$.

Step 1: Equation of the normal at $M$

With $\dot x=-a\sin t$, $\dot y=b\cos t$, the normal has slope $k_n=\dfrac{a\sin t}{b\cos t}$; clearing denominators and writing $c^{2}=a^{2}-b^{2}$, the normal is

\[ a\sin t\cdot x - b\cos t\cdot y - c^{2}\sin t\cos t = 0. \tag{1} \]

Step 2: Distances from the foci

The foci are $F_{1}=(c,0)$, $F_{2}=(-c,0)$. Using the point-to-line distance formula on (1): \[d_{1} = \frac{|ac\sin t - c^{2}\sin t\cos t|}{\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}} = \frac{c|\sin t|\,|a-c\cos t|}{\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}}, \tag{2}\]

\[ d_{2} = \frac{|-ac\sin t - c^{2}\sin t\cos t|}{\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}} = \frac{c|\sin t|\,|a+c\cos t|}{\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}}. \tag{3} \]

Step 3: The product

Since $|a-c\cos t||a+c\cos t| = a^{2}-c^{2}\cos^{2}t$, and substituting $c^{2}=a^{2}-b^{2}$, \[a^{2}-c^{2}\cos^{2}t = a^{2}(1-\cos^{2}t)+b^{2}\cos^{2}t = a^{2}\sin^{2}t+b^{2}\cos^{2}t, \tag{4}\]

which is exactly the denominator of $d_1d_2$, so it cancels: \[d_{1}d_{2} = \frac{c^{2}\sin^{2}t\,(a^{2}\sin^{2}t+b^{2}\cos^{2}t)}{a^{2}\sin^{2}t+b^{2}\cos^{2}t} = c^{2}\sin^{2}t. \tag{5}\]

Step 4: Verification

\[(a,b,t)=(3,2,0.7)\text{: } d_1d_2\approx2.07508,\ c^2\sin^2t\approx2.07508\ \checkmark\]

\[(a,b,t)=(5,1,1.2),\ (2,1.9,0.3)\text{: same check}\ \checkmark\]