A few properties for solving eclipse problems

For an eclipse centered at the origin x2a2+y2b2=1\dfrac { {x }^{ 2} }{ { a }^{ 2} } +\dfrac { {y }^{ 2} }{ { b}^{ 2} } =1:

  1. Any point PP on the eclipse can be expressed as (acosθ,bsinθ)(a\cos\theta ,b\sin\theta ) (or (asinθ,bcosθ)(a\sin\theta ,b\cos\theta ));

  2. Equation of the line tangent to the eclipse through point P is xcosθa+ysinθb=1\dfrac {x\cos\theta }{a } +\dfrac {y\sin\theta }{b } =1;

  3. The distance from the origin to the line ax+by=cax + by = c is ca2+b2\left|\dfrac { c }{ \sqrt { { a }^{2 }+{ b }^{ 2 } } } \right|;

image credit: jatin yadav.
#Geometry

Note by Leo Lei
5 years, 6 months ago

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Comments

Wow nice! You should put them in the wiki here!

Pi Han Goh - 5 years, 6 months ago
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