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Mathematics 20 Online
OpenStudy (anonymous):

find the equation such that the sum of the square of its distances from the two fixed points (a,0) and (-a,0) is constant

OpenStudy (anonymous):

\[2x^{2}/a+y^{2}/b =C\]

OpenStudy (anonymous):

its an elipse

OpenStudy (anonymous):

the definition itself of this curve is that the distance of every point on it is the sum of the distances from two point called foci

OpenStudy (anonymous):

Then just folow what the problem says: \[\sqrt{(x-a)^{2}+y ^{2}} + \sqrt{(x+a)^{2}+y ^{2}} = C\] and just rearange this a bit.

OpenStudy (anonymous):

|dw:1333819027791:dw| this is how more or less this could look. the slope is the tangent of the angle. so \[\alpha = arctg (-2/3)\] but you need angle with y axis, so it would be 90 - alpha

OpenStudy (anonymous):

sry, mistake, :) |dw:1333819247374:dw| but you need angle with y axis, so it would be 180 - alpha

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