Fancy Points on Ellipse

Let the ellipse have an equation of $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ , where $a > b$ . Then by rearranging the equation, any point on the ellipse can be expressed as $(k, \pm\frac{b}{a}\sqrt{a^2 - k^2})$ .

Using implicit differentiation on the equation of the ellipse, $\frac{2x}{a^2} + \frac{2y}{b^2}y' = 0$ , which solves to $y' = -\frac{b^2x}{a^2y}$ , so the slope of a normal line perpendicular to the slope of the tangent line is $\frac{a^2y}{b^2x}$ , and if it through an $x$ -coordinate of $k$ its slope is $\pm \frac{a\sqrt{a^2 - k^2}}{bk}$ and its equation is $y = \pm \frac{a\sqrt{a^2 - k^2}}{bk}(x - k) \pm \frac{b}{a}\sqrt{a^2 - k^2}$ or $y = \pm \frac{\sqrt{a^2 - k^2}}{abk}(a^2x + (a^2 - b^2)k)$ .

Without loss of generality (and to match the given picture), let the point with three distinct normals be point $P$ in the first quadrant with coordinates $(k_1, \frac{b}{a}\sqrt{a^2 - k_1^2})$ , let another point of normalcy be $Q$ in the fourth quadrant with coordinates $(k_2, -\frac{b}{a}\sqrt{a^2 - k_2^2})$ , and let the third point of normalcy be $R$ in the third quadrant with coordinates $(k_3, -\frac{b}{a}\sqrt{a^2 - k_3^2})$ . Also let $S$ be a point on the normal through $P$ not through $R$ or $Q$ .

Using the equations above, the slope through $PS$ is $m_1 = \frac{a\sqrt{a^2 - k_1^2}}{bk_1}$ , the slope through $PQ$ is $m_2 = -\frac{a\sqrt{a^2 - k_2^2}}{bk_2}$ , and the slope through $PR$ is $m_3 = -\frac{a\sqrt{a^2 - k_3^2}}{bk_3}$ , and the equations for $PQ$ and $PR$ are $y = -\frac{\sqrt{a^2 - k^2}}{abk}(a^2x + (a^2 - b^2)k)$ for $k = k_2$ and $k = k_3$ , respectively.

Since $P(k_1, \frac{b}{a}\sqrt{a^2 - k_1^2})$ is on both $PQ$ and $PR$ , we have:

$\frac{b}{a}\sqrt{a^2 - k_1^2} = -\frac{\sqrt{a^2 - k^2}}{abk}(a^2k_1 + (a^2 - b^2)k)$

which after rearranging simplifies to:

$k^4 - \frac{2a^2k_1}{a^2 - b^2}k^3 + \frac{(a^4 - b^4)k_1^2 - a^4(a^2 - 2b^2)}{(a^2 - b^2)^2}k^2 + \frac{2a^4k_1}{a^2 - b^2}k - \frac{a^6k_1^2}{(a^2 - b^2)^2} = 0$

where the roots of this equation represent the $x$ -coordinates of all the points of normalcy that go through $P$ . Being a quartic equation suggests that up to $4$ distinct normals can exist, but since we must have exactly $3$ distinct normals this suggests there is a double-root. Also, one of these points includes $P$ 's $x$ -coordinate of $k_1$ , so fortunately we can factor $(k - k_1)$ to obtain:

$(k - k_1)\bigg(k^3 - \frac{k_1(a^2 + b^2)}{a^2 - b^2}k^2 - \frac{a^4(a^2 - 2b^2)}{(a^2 - b^2)^2}k + \frac{k_1a^6}{(a^2 - b^2)^2}\bigg) = 0$

Without loss of generality, let the double-root be $k_2$ . Then $(k - k_1)(k - k_2)^2(k - k_3) = 0$ , or $(k - k_1)(k^3 - (2k_2 + k_3)k^2 + (k_2^2 + k_3)k - k_2^2k_3) = 0$ , and comparing coefficients with this equation to the quartic equation above gives a system of equations where:

$2k_2 + k_3 = \frac{k_1(a^2 + b^2)}{a^2 - b^2}, k_2^2 + k_3 = -\frac{a^4(a^2 - 2b^2)}{(a^2 - b^2)^2}, k_2^2k_3 = -\frac{k_1a^6}{(a^2 - b^2)^2}$

which solves to:

$k_1 = \sqrt{\frac{a^4(a^2 - 2b^2)^3}{(a^2 - b^2)(a^2 + b^2)^3}}, k_2 = \sqrt{\frac{a^4(a^2 - 2b^2)}{(a^2 - b^2)(a^2 + b^2)}}, k_3 = -\sqrt{\frac{a^8(a^2 - 2b^2)}{(a^2 - b^2)^3(a^2 + b^2)}}$

for $k_1 > 0$ , $k_2 > 0$ , and $k_3 < 0$ . Plugging these values into $m_1 = \frac{a\sqrt{a^2 - k_1^2}}{bk_1}$ , $m_2 = \frac{a\sqrt{a^2 - k_2^2}}{bk_2}$ , and $m_3 = \frac{a\sqrt{a^2 - k_3^2}}{bk_3}$ as defined from above gives:

$m_1 = \sqrt{\frac{(2a^2 - b^2)^3}{(a^2 - 2b^2)^3}}, m_2 = -\sqrt{\frac{2a^2 - b^2}{a^2 - 2b^2}}, m_3 = \sqrt{\frac{b^4(2a^2 - b^2)}{a^4(a^2 - 2b^2)}}$

Using the angle between two lines equation gives $\tan \theta_1 = \frac{m_2 - m_1}{1 + m_1m_2} = \frac{\sqrt{(a^2 - 2b^2)(2a^2 - b^2)}}{a^2 + b^2}$ and $\tan \theta_2 = \frac{m_1 - m_3}{1 + m_1m_3} = \frac{2\sqrt{(a^2 - 2b^2)(2a^2 - b^2)}}{a^2 + b^2} = 2\tan \theta_1$ .

Therefore, $\frac{\tan \theta_1}{\tan \theta_2} = \frac{\tan \theta_1}{2 \tan \theta_1} = \frac{1}{2} = \boxed{0.5}$ .