Find the length of the segment that the ellipse x=a cos

Find the length of the segment that the ellipse $x=a\cos t,\ y=b\sin t$ ($a>b>0$) cuts on its own normal at an arbitrary point $M$.

Step 1: Parametrise the normal line

The normal direction at $M=(a\cos t,b\sin t)$ (perpendicular to the tangent direction $(-a\sin t,b\cos t)$) is $(b\cos t,\,a\sin t)$, so points on the normal are

\[ x = a\cos t + b\cos t\cdot s, \qquad y = b\sin t + a\sin t\cdot s. \tag{1} \]

Step 2: Substitute into the ellipse equation

Substituting (1) into $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, the constant terms cancel (since $\cos^2t+\sin^2t=1$), leaving a linear factor of $s$ times an affine expression in $s$; besides the root $s=0$ (the point $M$ itself), the second intersection is at

\[ s_{1} = -\frac{\dfrac{2b\cos^{2}t}{a}+\dfrac{2a\sin^{2}t}{b}}{\dfrac{b^{2}\cos^{2}t}{a^{2}}+\dfrac{a^{2}\sin^{2}t}{b^{2}}} = -\frac{2ab\bigl(b^{2}\cos^{2}t+a^{2}\sin^{2}t\bigr)}{b^{4}\cos^{2}t+a^{4}\sin^{2}t}. \tag{2} \]

Step 3: Chord length

The direction vector $(b\cos t,a\sin t)$ has length $\ell=\sqrt{b^{2}\cos^{2}t+a^{2}\sin^{2}t}$, so with $D=a^{2}\sin^{2}t+b^{2}\cos^{2}t$: \[|MN| = |s_{1}|\cdot\ell = \frac{2ab\,D^{3/2}}{a^{4}\sin^{2}t+b^{4}\cos^{2}t}. \tag{3}\]

Step 4: Special-case checks

At $t=0$ (vertex $(a,0)$): $D=b^2$, denominator $b^4$, giving $|MN|=2a$ — correct, since the normal there is the major axis itself. At $t=\pi/2$ ($(0,b)$): $D=a^2$, denominator $a^4$, giving $|MN|=2b$ — the normal is the minor axis.

Step 5: Verification

\[t=0\text{: formula gives } 2a\ \checkmark;\qquad t=\pi/2\text{: formula gives } 2b\ \checkmark\]

\[(a,b,t)=(3,2,0.7)\text{: chord found by direct intersection} \approx4.18103,\ \text{formula} \approx4.18103\ \checkmark\]

\[(a,b,t)=(5,1,1.2),\ (4,2.5,0.9)\text{: same check}\ \checkmark\]