Angle of view

The set of all points $p$ where two tangents are perpendicular for any conic is an orthoptic , and the orthoptic for an ellipse in the form of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is a circle with the equation $x^2 + y^2 = a^2 + b^2$ .

Proof: Using implicit differentiation on $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ gives $\frac{2x}{a^2} + \frac{2y}{b^2}y' = 0$ , which simplifies to $y' = -\frac{b^2 x}{a^2 y}$ , and is the slope of the curve $m = y'$ at any point $(x, y)$ . Let $(p, q)$ and $(r, s)$ be points on the ellipse such that their tangent are perpendicular. Then the equation of the tangent line at a point $(p, q)$ is $y - q = -\frac{b^2 p}{a^2 q}(x - p)$ , and the equation of the tangent line at a point $(r, s)$ is $y - s = -\frac{b^2 r}{a^2 s}(x - r)$ . Now, if $(p, q)$ and $(r, s)$ have tangents that are perpendicular to each other, $m_1m_2 = -1$ , or $(-\frac{b^2 p}{a^2 q})(-\frac{b^2 r}{a^2 s}) = -1$ . Using this and the fact that $(r, s)$ is on the ellipse so $\frac{r^2}{a^2 } + \frac{s^2}{b^2 } = 1$ gives $r = \pm \frac{a^4q}{\sqrt{b^6p^2 + a^6q^2}}$ and $s = \mp \frac{b^4p}{\sqrt{b^6p^2 + a^6q^2}}$ . Solving the two tangent line equations for $x$ and $y$ gives $x = \frac{\pm a^2 q\sqrt{b^6p^2 + a^6q^2} + b^4 a^2 p}{b^4p^2 + a^4q^2}$ and $y = \frac{\mp b^2 p\sqrt{b^6p^2 + a^6q^2} + a^4 b^2 q}{b^4p^2 + a^4q^2}$ , and $x^2 + y^2$ simplifies to $a^2 + b^2$ .