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Question Number 46737 by peter frank last updated on 30/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18
	$Q(asinθ_1 ,bcosθ_1 ) R(asinθ_2 ,bcosθ_2 ) mid point QR is P P=((a(sinθ_1 +sinθ_2 ))/2),((b(cosθ_1 +cosθ_2 ))/2) ={a.sin(((θ_1 +θ_2 )/2))cos(((θ_1 −θ_2 )/2)) ,bcos(((θ_1 +θ_2 )/2))cos(((θ_1 −θ_2 )/2)} α=asin(((θ_1 +θ_2 )/2))cos(((θ_1 −θ_2 )/2)) β=bcos(((θ_1 +θ_2 )/2))cos(((θ_1 −θ_2 )/2)) (∝/β)=(a/b)tan(((θ_1 +θ_2 )/2)) now the eqn of st line ((y−bcosθ_1 )/(b(cosθ_2 −cosθ_1 )))=((x−asinθ_1 )/(a(sinθ_2 −sinθ_1 ))) gradient m=((b(cosθ_2 −cosθ_1 ))/(a(sinθ_2 −sinθ_1 )))=(b/a)×((2sin(((θ_1 +θ_2 )/2))sin(((θ_1 −θ_2 )/2)))/(2cos(((θ_1 +θ_2 )/2))sin(((θ_2 −θ_1 )/2)))) m=−(b/a)×tan(((θ_1 +θ_2 )/2)) tan(((θ_1 +θ_2 )/2))=((−ma)/b) [m=constant gradient] it is already derived (α/β)=(a/b)tan(((θ_1 +θ_2 )/2)) (α/β)=(a/b)×((−ma)/b) so locus is (x/y)=((−ma^2 )/b^2 ) ((b^2 x)/(a^2 y))+m=0 [$
	$Q (a s i n θ_{1}, b c o s θ_{1})$ $R (a s i n θ_{2}, b c o s θ_{2})$ $m i d p o i n t Q R i s P$ $P = \frac{a (s i n θ_{1} + s i n θ_{2})}{2}, \frac{b (c o s θ_{1} + c o s θ_{2})}{2}$ $= {a . s i n (\frac{θ_{1} + θ_{2}}{2}) c o s (\frac{θ_{1} - θ_{2}}{2}), b c o s (\frac{θ_{1} + θ_{2}}{2}) c o s (\frac{θ_{1} - θ_{2}}{2}}$ $α = a s i n (\frac{θ_{1} + θ_{2}}{2}) c o s (\frac{θ_{1} - θ_{2}}{2})$ $β = b c o s (\frac{θ_{1} + θ_{2}}{2}) c o s (\frac{θ_{1} - θ_{2}}{2})$ $\frac{\propto}{β} = \frac{a}{b} t a n (\frac{θ_{1} + θ_{2}}{2})$ $n o w t h e e q n o f s t l i n e$ $\frac{y - b c o s θ_{1}}{b (c o s θ_{2} - c o s θ_{1})} = \frac{x - a s i n θ_{1}}{a (s i n θ_{2} - s i n θ_{1})}$ $g r a d i e n t m = \frac{b (c o s θ_{2} - c o s θ_{1})}{a (s i n θ_{2} - s i n θ_{1})} = \frac{b}{a} \times \frac{2 s i n (\frac{θ_{1} + θ_{2}}{2}) s i n (\frac{θ_{1} - θ_{2}}{2})}{2 c o s (\frac{θ_{1} + θ_{2}}{2}) s i n (\frac{θ_{2} - θ_{1}}{2})}$ $m = - \frac{b}{a} \times t a n (\frac{θ_{1} + θ_{2}}{2})$ $t a n (\frac{θ_{1} + θ_{2}}{2}) = \frac{- m a}{b} [m = c o n s t a n t g r a d i e n t]$ $i t i s a l r e a d y d e r i v e d$ $\frac{α}{β} = \frac{a}{b} t a n (\frac{θ_{1} + θ_{2}}{2})$ $\frac{α}{β} = \frac{a}{b} \times \frac{- m a}{b}$ $s o l o c u s i s$ $\frac{x}{y} = \frac{- {m a}^{2}}{b^{2}}$ $\frac{b^{2} x}{a^{2} y} + m = 0 [$
	Commented by peter frank last updated on 31/Oct/18

	$thank you$
	Answered by MrW3 last updated on 31/Oct/18

	$Q (x_{1}, y_{1}) a n d R (x_{2}, y_{2})$ $\frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} = 1 . . . (i)$ $\frac{x_{2}^{2}}{a^{2}} + \frac{y_{2}^{2}}{b^{2}} = 1 . . . (i i)$ $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = m . . . (i i i)$ $P (u, v) w i t h u = \frac{x_{1} + x_{2}}{2}, v = \frac{y_{1} + y_{2}}{2}$ $(i i) - (i) :$ $\frac{(x_{1} + x_{2}) (x_{2} - x_{1})}{a^{2}} + \frac{(y_{1} + y_{2}) (y_{2} - y_{1})}{b^{2}} = 0$ $\frac{(x_{1} + x_{2})}{2 a^{2}} + \frac{(y_{1} + y_{2}) (y_{2} - y_{1})}{2 b^{2} (x_{2} - x_{1})} = 0$ $\frac{u}{a^{2}} + \frac{v m}{b^{2}} = 0$ $o r$ $\frac{x}{a^{2}} + \frac{m y}{b^{2}} = 0 o r y = - \frac{b^{2}}{{m a}^{2}} x$ $i . e . t h e l o c u s o f p o i n t P i s a s e c t i o n o f l i n e .$
	Commented by peter frank last updated on 31/Oct/18

	$thank you$
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