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Geometry Level pending

In an orthonormal system ( O ; u ; v ) (O; \vec u; \vec v ) , consider M ( x , y ) M(x,y) moving on ( E ) (E) : 25 ( x 2 + y 2 ) = ( 3 x 16 ) 2 25(x^2+y^2) = (3x-16)^2 having a directrix ( D ) (D) : x = 16 3 x = \dfrac{16}3 . Let ( u ; O M ) = θ ( \vec u ; \vec{OM} ) = \theta . ( O M ) (OM) meets ( E ) (E) at M M' .

Given that 1 O M + 1 O M = a b \dfrac1{OM} + \dfrac1{OM'} = \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 13.

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1 solution

Mohammad Hamdar
Apr 3, 2016

T h e r e d u c e d e q u a t i o n o f ( E ) i s ( x + 3 ) 2 25 + y 2 16 = 1 i t s e c c e n t r i c i t y i s e = 3 5 . O i s a f o c u s o f ( E ) . z M = r ( cos θ + i sin θ ) , M ( O M cos θ ; O M sin θ ) . d = d ( M ; ( D ) ) = 3 O M cos θ 16 3 = 16 3 O M cos θ 3 s i n c e 0 < x M < 2 , 16 < 3 x M 16 < 10 O M d = e = 3 5 O M = 3 5 d ; 5 O M = 16 3 O M cos θ O M = 16 5 + 3 cos θ O M = 16 5 + 3 cos ( θ π ) = 16 5 3 cos θ 1 O M + 1 O M = 5 + 3 cos θ 16 + 5 3 cos θ 16 = 5 8 The\quad reduced\quad equation\quad of\quad (E)\quad is\quad \frac { { (x+3) }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 16 } =1\\ its\quad eccentricity\quad is\quad e=\frac { 3 }{ 5 } .\quad O\quad is\quad a\quad focus\quad of\quad (E).\\ { z }_{ M }=r(\cos { \theta } +i\sin { \theta } ),\quad M(OM\cos { \theta ;OM\sin { \theta } ) } .\\ d=d(M;(D))=\frac { |3OM\cos { \theta } -16| }{ 3 } =\frac { 16-3OM\cos { \theta } }{ 3 } \\ since\quad 0<{ x }_{ M }<2\quad ,\quad -16<3{ x }_{ M }-16<-10\\ \frac { OM }{ d } =e=\frac { 3 }{ 5 } \Rightarrow OM=\frac { 3 }{ 5 } d;\quad 5OM=16-3OM\cos { \theta } \\ \quad \quad \quad \quad \quad OM=\frac { 16 }{ 5+3\cos { \theta } } \\ OM'=\frac { 16 }{ 5+3\cos { (\theta -\pi ) } } =\frac { 16 }{ 5-3\cos { \theta } } \\ \frac { 1 }{ OM } +\frac { 1 }{ OM' } =\frac { 5+3\cos { \theta } }{ 16 } +\frac { 5-3\cos { \theta } }{ 16 } =\frac { 5 }{ 8 }

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