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Question Number 46829 by peter frank last updated on 01/Nov/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
	$eqn of focal chord is y=tanθ(x−(√(a^2 −b^2 )) ) focaus((√(a^2 −b^2 )) ,0) length of focal chord (√((x_2 −x_2 )^2 +(y_2 −y_1 )^2 )) solve y=tanθ(x−(√(a^2 −b^2 )) )and (x^2 /a^2 )+(y^2 /b^2 )=1 to get (x_1 ,y_1 ) and(x_2 ,y_2 ) y=tanθ(x−p) (x^2 /a^2 )+((tan^2 θ(x−p)^2 )/b^2 )=1 (p=(√(a^2 −b^2 )) ) b^2 x^2 +a^2 tan^2 θ((x^2 −2xp+p^2 )=a^2 b^2 x^2 (b^2 +a^2 tan^2 θ)−x(2p×a^2 tan^2 θ)+a^2 tan^2 θp^2 −a^2 b^2 =0 x_1 +x_2 =((2p×a^2 tan^2 θ)/((b^2 +a^2 tan^2 θ)))←x_1 +x_2 x_1 x_2 =((a^2 p^2 tan^2 θ−a^2 b^2 )/(b^2 +a^2 tan^2 θ))←x_1 x_2 (x_2 −x_1 )^2 =(x_2 +x_1 )^2 −4x_1 x_2 =((4p^2 a^4 tan^4 θ)/((b^2 +a^2 tan^2 θ)^2 ))−((4a^2 p^2 tan^2 θ−4a^2 b^2 )/(b^2 +a^2 tan^2 θ)) =((4p^2 a^4 tan^4 θ−(b^2 +a^2 tan^2 θ)(4a^2 p^2 tan^2 θ−4a^2 b^2 ))/((b^2 +a^2 tan^2 θ)^2 )) =((4p^2 a^4 tan^4 θ−4a^2 b^2 p^2 tan^2 θ+4a^2 b^4 −4a^4 p^2 tan^4 θ+4a^4 b^2 tan^2 θ)/((b^2 +a^2 tan^2 θ)^2 )) =((4a^2 b^4 −4a^2 b^2 p^2 tan^2 θ+4a^4 b^2 tan^2 θ)/((b^2 +a^2 tan^2 θ)^2 ))←(x_2 −x_1 )^2 =((4a^2 b^4 +4a^2 b^2 tan^2 θ(a^2 −p^2 ))/((b^2 +a^2 tan^2 θ)^2 )) =((4a^2 b^4 +4a^2 b^4 tan^2 θ)/D_r ^2 ) [p^2 =a^2 −b^2 ] =((4a^2 b^4 (1+tan^2 θ))/D_r ^2 )=((4a^2 b^4 sec^2 θ)/D_r ^2 ) so (x_2 −x_1 )=((2ab^2 secθ)/D_r ) y=(x−p)tanθ y_2 −y_1 =tanθ(x_2 −p−x_1 +p) y_2 −y_1 =tanθ(x_2 −x_1 ) {(y_2 −y_1 )^2 +(x_2 −x_1 )^2 }^(1/2) =[(x_2 −x_1 )^2 {tan^2 θ+1}]^(1/2) =(x_2 −x_1 )secθ =((2ab^2 sec^2 θ)/D_r )=((2ab^2 )/(cos^2 θ[b^2 +a^2 tan^2 θ])) =((2ab^2 )/(b^2 cos^2 θ+a^2 sin^2 θ)) ultimately got the answer$
	$e q n o f f o c a l c h o r d i s$ $y = t a n θ (x - \sqrt{a^{2} - b^{2}}) f o c a u s (\sqrt{a^{2} - b^{2}}, 0)$ $l e n g t h o f f o c a l c h o r d$ $\sqrt{{(x_{2} - x_{2})}^{2} + {(y_{2} - y_{1})}^{2}}$ $s o l v e y = t a n θ (x - \sqrt{a^{2} - b^{2}}) a n d \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $t o g e t (x_{1}, y_{1}) a n d (x_{2}, y_{2})$ $y = t a n θ (x - p)$ $\frac{x^{2}}{a^{2}} + \frac{{t a n}^{2} θ {(x - p)}^{2}}{b^{2}} = 1 (p = \sqrt{a^{2} - b^{2}})$ $b^{2} x^{2} + a^{2} {t a n}^{2} θ ((x^{2} - 2 x p + p^{2}) = a^{2} b^{2}$ $x^{2} (b^{2} + a^{2} {t a n}^{2} θ) - x (2 p \times a^{2} {t a n}^{2} θ) + a^{2} {t a n}^{2} θ p^{2} - a^{2} b^{2} = 0$ $x_{1} + x_{2} = \frac{2 p \times a^{2} {t a n}^{2} θ}{(b^{2} + a^{2} {t a n}^{2} θ}) \leftarrow x_{1} + x_{2}$ $x_{1} x_{2} = \frac{a^{2} p^{2} {t a n}^{2} θ - a^{2} b^{2}}{b^{2} + a^{2} {t a n}^{2} θ} \leftarrow x_{1} x_{2}$ ${(x_{2} - x_{1})}^{2} = {(x_{2} + x_{1})}^{2} - 4 x_{1} x_{2}$ $= \frac{4 p^{2} a^{4} {t a n}^{4} θ}{{(b^{2} + a^{2} {t a n}^{2} θ)}^{2}} - \frac{4 a^{2} p^{2} {t a n}^{2} θ - 4 a^{2} b^{2}}{b^{2} + a^{2} {t a n}^{2} θ}$ $= \frac{4 p^{2} a^{4} {t a n}^{4} θ - (b^{2} + a^{2} {t a n}^{2} θ) (4 a^{2} p^{2} {t a n}^{2} θ - 4 a^{2} b^{2})}{{(b^{2} + a^{2} {t a n}^{2} θ)}^{2}}$ $= \frac{4 p^{2} a^{4} {t a n}^{4} θ - 4 a^{2} b^{2} p^{2} {t a n}^{2} θ + 4 a^{2} b^{4} - 4 a^{4} p^{2} {t a n}^{4} θ + 4 a^{4} b^{2} {t a n}^{2} θ}{{(b^{2} + a^{2} {t a n}^{2} θ)}^{2}}$ $= \frac{4 a^{2} b^{4} - 4 a^{2} b^{2} p^{2} {t a n}^{2} θ + 4 a^{4} b^{2} {t a n}^{2} θ}{{(b^{2} + a^{2} {t a n}^{2} θ)}^{2}} \leftarrow {(x_{2} - x_{1})}^{2}$ $= \frac{4 a^{2} b^{4} + 4 a^{2} b^{2} {t a n}^{2} θ (a^{2} - p^{2})}{{(b^{2} + a^{2} {t a n}^{2} θ)}^{2}}$ $= \frac{4 a^{2} b^{4} + 4 a^{2} b^{4} {t a n}^{2} θ}{D_{r}^{2}} [p^{2} = a^{2} - b^{2}]$ $= \frac{4 a^{2} b^{4} (1 + {t a n}^{2} θ)}{D_{r}^{2}} = \frac{4 a^{2} b^{4} {s e c}^{2} θ}{D_{r}^{2}}$ $s o (x_{2} - x_{1}) = \frac{2 {a b}^{2} s e c θ}{D_{r}}$ $y = (x - p) t a n θ$ $y_{2} - y_{1} = t a n θ (x_{2} - p - x_{1} + p)$ $y_{2} - y_{1} = t a n θ (x_{2} - x_{1})$ ${{(y_{2} - y_{1})}^{2} + {(x_{2} - x_{1})}^{2}}^{\frac{1}{2}}$ $= {[{(x_{2} - x_{1})}^{2} {{t a n}^{2} θ + 1}]}^{\frac{1}{2}}$ $= (x_{2} - x_{1}) s e c θ$ $= \frac{2 {a b}^{2} {s e c}^{2} θ}{D_{r}} = \frac{2 {a b}^{2}}{{c o s}^{2} θ [b^{2} + a^{2} {t a n}^{2} θ]}$ $= \frac{2 {a b}^{2}}{b^{2} {c o s}^{2} θ + a^{2} {s i n}^{2} θ}$ $u l t i m a t e l y g o t t h e a n s w e r$
	Commented by peter frank last updated on 01/Nov/18

	$God bless you sir . thank you$
	Commented by peter frank last updated on 01/Nov/18

	$D_{r} mean ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18

	$D_{r} = d e n o m i n a t o r . . . i h a v e p l a c e d D_{r} t o m i m i s e$ $t i m e . . . n o t w r i t i n g t h e v a l u e o f D_{r} a g a i n a n a a g a i n$
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