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

Any point \(P\) on the eclipse can be expressed as \((a\cos\theta ,b\sin\theta )\) (or \((a\sin\theta ,b\cos\theta )\));

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

The distance from the origin to the line \(ax + by = c \) is \(\left|\dfrac { c }{ \sqrt { { a }^{2 }+{ b }^{ 2 } } } \right|\);

## Comments

Sort by:

TopNewestWow nice! You should put them in the wiki here! – Pi Han Goh · 9 months, 2 weeks ago

Log in to reply