Ellipse properties

Definitions shown in Ellipse

Ellipse and tangent

Let an ellipse $\theta$ with foci $F, F'$ and the point $S \in FF'$ outside $\theta$ be given. Let $\Omega$ be the circumcircle of given ellipse. Let $S'$ be the inverse of a point $S$ with respect to $\Omega.$ Let $C$ be the point of the ellipse such that $CS' \perp FF'.$ Prove that $CS$ is tangent to the ellipse.

Proof

Let equation of the ellipse be $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \implies \frac {x}{a^2} + \frac {yy'}{b^2} = 0.$

The radus of the circumcircle is $a,$ so $SO \cdot S'O = a^2.$ Denote $y = CS', x = OS',$ so the point $C(x,y).$ The slope of the tangent at point $C$ is: $y' = - \frac{xb^2}{ya^2} = - \frac{xyb^2}{y^2 a^2} = - \frac{xyb^2}{a^2 b^2 - x^2 b^2}=$ $= - \frac{xy}{a^2 - x^2}= - \frac{y}{\frac {a^2}{x} - x} = \frac{y}{x - SO} = \frac{CS'}{SS'}. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

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Ellipse properties

Ellipse and tangent