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ilikebigbutts ilikebigbutts
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12 years ago
Show by implicit differentiation that the tangent to the ellipse

(x^2/a^2) + (y^2/b^2) = 1

at the point (c,d) is

(cx/a^2) + (dy/b^2) = 1
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wrote...
#1
12 years ago
2x/a² + 2y/b² y' = 0
y' = (b²/a²) (-x/y)
= (b²/a²)(-c/d) .... at the point (c,d)

y - d = (b²/a²) (-c/d) (x - c) ..... is the tangent line: (y - y?) = m (x - x?)
dy - d² = (b²/a²) (c² - cx)
dy/b² - d²/b² = c²/a² - cx/a²
cx/a² + dy/b² = c²/a² + d²/b²
cx/a² + dy/b² = 1 .... note that if (c,d) is on the ellipse, then c²/a² + d²/b² = 1

Answer: see above
wrote...
#2
12 years ago
See:
(x^2/a^2) + (y^2/b^2) = 1
(2x/a^2) + (2y.y'/b^2) = 0
(2y.y'/b^2) = -(2x/a^2)
2y.y' = -(2xb^2/a^2)
y' = -(2xb^2/2ya^2)
y' = -xb^2/ya^2
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