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Difference between revisions of "Ellipse properties"

(Created page with "Definitions shown in Ellipse ==Ellipse and tangent== Let an ellipse <math>\theta</math> with foci <math>F, F'</math> and the point <math>S \in FF'</math> outside <math>\...")
 
(→‎Ellipse and tangent)
 
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==Ellipse and tangent==  
 
==Ellipse and tangent==  
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Let an ellipse <math>\theta</math> with foci <math>F, F'</math> and the point <math>S \in FF'</math> outside <math>\theta</math> be given.  
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[[File:Ellipse tagent.png|350px|right]]
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Let <math>\Omega</math> be the circumcircle of given ellipse.
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Let an ellipse <math>\theta</math> with foci <math>F, F'</math> and the point <math>S \in FF'</math> outside <math>\theta</math> be given.
 +
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Let <math>\Omega</math> centered at <math>O</math> be the circumcircle of given ellipse.
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Let <math>S'</math> be the inverse of a point <math>S</math> with respect to <math>\Omega.</math>
 
Let <math>S'</math> be the inverse of a point <math>S</math> with respect to <math>\Omega.</math>
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Let <math>C</math> be the point of the ellipse such that <math>CS' \perp FF'.</math>
 
Let <math>C</math> be the point of the ellipse such that <math>CS' \perp FF'.</math>
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Prove that <math>CS</math> is tangent to the ellipse.
 
Prove that <math>CS</math> is tangent to the ellipse.
  
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
  
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Let equation of the ellipse be <math>\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \implies \frac {x}{a^2} + \frac {yy'}{b^2} = 0.</math>
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Let equation of the ellipse be <cmath>\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \implies \frac {x}{a^2} + \frac {yy'}{b^2} = 0.</cmath>
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The radus of the circumcircle <math>\Omega</math> is <math>a,</math> so <math>SO \cdot S'O = a^2.</math>
  
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The radus of the circumcircle is <math>a,</math> so <math>SO \cdot S'O = a^2.</math>
 
 
Denote <math>y = CS', x = OS',</math> so the point  <math>C(x,y).</math>
 
Denote <math>y = CS', x = OS',</math> so the point  <math>C(x,y).</math>
 
The slope of the tangent at point <math>C</math> is:
 
The slope of the tangent at point <math>C</math> is:

Latest revision as of 14:28, 12 December 2024

Definitions shown in Ellipse

Ellipse and tangent

Ellipse tagent.png

Let an ellipse $\theta$ with foci $F, F'$ and the point $S \in FF'$ outside $\theta$ be given.

Let $\Omega$ centered at $O$ 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 $\Omega$ 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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