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March 27, 2015

March 27, 2015

Posted by **Anonymous** on Tuesday, November 15, 2011 at 4:59pm.

- calculus -
**Steve**, Tuesday, November 15, 2011 at 6:28pmBy inspection, the curves intersect at (1,1) and (1,-1)

ellipse: 4x + 6yy' = 0

y' = -2x/3y

at (1,1) slope = -2/3

at (1,-1) slope = 2/3

semicubical parabola: 2yy' = 3x^2

y' = 3x/2y

at (1,1) slope = 3/2

at (1,-1) slope = -3/2

The slopes at the intersections are negative reciprocals; hence the curves are orthogonal.

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