The escentric angle of a point lying in the first quadrant of ellipse x2/a2+y2/b2=1 be α and the line joining the centre to the point makes an angle β with x-axis then α-β will be maximum when α =??
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The point P will have coordinates (acosα, bsinα) and so tanβ =(b/a)tanα
Use the formula tan(A-B)=[tanA - tanB]/[1+tanAtanB} and you will get
tan(α-β) = (a-b)tanα/(a+btan^2(α))
This is max when t/(a+bt^2) is max, where t=tanα
This can be found from where the derivative is zero,
and you find t=sqrt(a/b), α= tan^-1(sqrt(a/b))
You should check that this does give a max and not a min.
Note that 0<α<=2pi.
Use the formula tan(A-B)=[tanA - tanB]/[1+tanAtanB} and you will get
tan(α-β) = (a-b)tanα/(a+btan^2(α))
This is max when t/(a+bt^2) is max, where t=tanα
This can be found from where the derivative is zero,
and you find t=sqrt(a/b), α= tan^-1(sqrt(a/b))
You should check that this does give a max and not a min.
Note that 0<α<=2pi.