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Question Number 103228 by ajfour last updated on 13/Jul/20

Commented by ajfour last updated on 13/Jul/20

$C a l c u l a t e h, k i n t e r m s o f e l l i p s e$ $p a r a m e t e r s a, b .$
	Answered by mr W last updated on 13/Jul/20

	Commented by mr W last updated on 15/Jul/20

	$l e t μ = \frac{b}{a}$ $C (a \cos θ, b \sin θ)$ $y^{'} = - \frac{b \cos θ}{a \sin θ} = - \frac{μ}{\tan θ}$ $A (a \cos φ, b \sin φ)$ $y^{'} = - \frac{b \cos φ}{a \sin φ} = - \frac{μ}{\tan φ}$ $(- \frac{μ}{\tan φ}) (- \frac{μ}{\tan θ}) = - 1$ $\tan φ = - \frac{μ^{2}}{\tan θ}$ $\Rightarrow φ = π - \tan^{- 1} (\frac{μ^{2}}{\tan θ})$ $e q n . o f C B :$ $y = \frac{\tan θ}{μ} (x - a \cos θ) + b \sin θ$ $e q n . o f A B :$ $y = \frac{\tan φ}{μ} (x - a \cos φ) + b \sin φ$ $\frac{\tan φ}{μ} x_{B} + a (μ - \frac{1}{μ}) \sin φ = \frac{\tan θ}{μ} x_{B} + a (μ - \frac{1}{μ}) \sin θ$ $\Rightarrow \frac{x_{B}}{a} = \frac{(1 - μ^{2}) (\sin θ - \sin φ)}{\tan θ - \tan φ}$ $\Rightarrow \frac{y_{B}}{b} = (\frac{1 - μ^{2}}{μ^{2}}) [\frac{(\sin θ - \sin φ) \tan θ}{\tan θ - \tan φ} - \sin θ]$ $\frac{x_{B}^{2}}{a^{2}} + \frac{y_{B}^{2}}{b^{2}} = 1$ ${[\frac{μ^{2} (\sin θ - \sin φ)}{\tan θ - \tan φ}]}^{2} + {[\frac{(\sin θ - \sin φ) \tan θ}{\tan θ - \tan φ} - \sin θ]}^{2} = {(\frac{μ^{2}}{1 - μ^{2}})}^{2}$ $\Rightarrow \tan^{2} θ {(\sin θ \sqrt{μ^{4} + \tan^{2} θ} - μ^{2})}^{2}$ $+ {(\sin θ \sqrt{μ^{4} + \tan^{2} θ} + \tan^{2} θ)}^{2}$ $= (μ^{4} + \tan^{2} θ) {(\frac{μ^{2} + \tan^{2} θ}{1 - μ^{2}})}^{2}$ $\Rightarrow θ = . . .$ $h = C B = \sqrt{{(x_{B} - a \cos θ)}^{2} + {(y_{B} - b \sin θ)}^{2}}$ $k = A B = \sqrt{{(x_{B} - a \cos φ)}^{2} + {(y_{B} - b \sin φ)}^{2}}$
	Commented by mr W last updated on 13/Jul/20

	Commented by mr W last updated on 13/Jul/20

	Commented by mr W last updated on 14/Jul/20

	Commented by ajfour last updated on 17/Jul/20
	$I have a nice way that results in quite a failure...please check n help! Let B(p,q) & inclination of BC is θ ⇒ inclination of AB is 90°+θ. BC = h, AB=k . B is on ellipse ⇒ b^2 p^2 +a^2 q^2 =a^2 b^2 _(−−−−−−−−−−−) ^(−−−−−−−−−−−) x_C = p+hcos θ ; y_C = q+hsin θ x_A = p−ksin θ ; y_A = q+kcos θ Ellipse eq: (x^2 /a^2 )+(y^2 /b^2 )=1 ⇒ slope of a normal at (x,y) is m = −(dx/dy)= ((a^2 /b^2 ))((y/x)) m_(BC) = tan θ = ((a^2 /b^2 ))(((q+hsin θ)/(p+hcos θ))) ..(i) m_(AB) =−(1/(tan θ))=((a^2 /b^2 ))(((q+kcos θ)/(p−ksin θ))) ..(ii) ⇒ (((b^2 p)/(cos θ))−((a^2 q)/(sin θ)))=(a^2 −b^2 )h ...(I) ⇒ −(((b^2 p)/(sin θ))+((a^2 q)/(cos θ)))=(a^2 −b^2 )k ...(II) Now C and A lie on ellipse ⇒ (x_C ^2 /a^2 )+(y_C ^2 /b^2 )=1 or b^2 (x_C )^2 +a^2 (y_C )^2 =a^2 b^2 ⇒b^2 (p+hcos θ)^2 +a^2 (q+hsin θ)^2 =a^2 b^2 and as b^2 p^2 +a^2 q^2 =a^2 b^2 so we have 2h(b^2 pcos θ+a^2 qsin θ) +h^2 (a^2 sin^2 θ+b^2 cos^2 θ) = 0 ⇒ −(((b^2 p)/(sin θ))+((a^2 q)/(sin θ)))=((h(a^2 sin^2 θ+b^2 cos^2 θ))/(2sin θcos θ)) Now using (II), (a^2 −b^2 )k=((h(a^2 sin^2 θ+b^2 cos^2 θ))/(2sin θcos θ)) ...(iii) (x_A ^2 /a^2 )+(y_A ^2 /b^2 )=1 ⇒ b^2 (x_A )^2 +a^2 (y_A )^2 =a^2 b^2 ⇒ b^2 (p−ksin θ)^2 +a^2 (q+kcos θ)^2 =a^2 b^2 and as b^2 p^2 +a^2 q^2 =a^2 b^2 so we have 2k(−b^2 psin θ+a^2 qcos θ) +k^2 (b^2 sin^2 θ+a^2 cos^2 θ) = 0 ⇒ ((b^2 p)/(cos θ))−((a^2 q)/(sin θ)) = ((k(b^2 sin^2 θ+a^2 cos^2 θ))/(2sin θcos θ)) Now using (I), (a^2 −b^2 )h=((k(b^2 sin^2 θ+a^2 cos^2 θ))/(2sin θcos θ)) ...(iv) and already we have (a^2 −b^2 )k=((h(a^2 sin^2 θ+b^2 cos^2 θ))/(2sin θcos θ)) ...(iii) Now (iii)×(iv) with m=tan θ gives 4(a^2 −b^2 )^2 m^2 =(b^2 m^2 +a^2 )(a^2 m^2 +b^3 ) ⇒ m^4 +{((a^4 +b^4 −4(a^2 −b^2 )^2 )/(a^2 b^2 ))}m^2 +1=0 let a/b = μ ⇒ m^4 +{μ^2 +(1/μ^2 )−4(μ−(1/μ))^2 }m^2 +1=0 {m^2 +1−(3/2)(μ−(1/μ))^2 }^2 = [1−(3/2)(μ−(1/μ))^2 ]^2 −1 Now tragedy is unless 𝛍=1 , m^2 ∉R so mrW Sir kindly help me out of this turmoil .... If m=tan θ, were real h and k could easily be obtained...$
	$I h a v e a n i c e w a y t h a t r e s u l t s i n$ $q u i t e a f a i l u r e . . . p l e a s e c h e c k n h e l p!$ $L e t$ $B (p, q) & i n c l i n a t i o n o f B C i s θ$ $\Rightarrow i n c l i n a t i o n o f A B i s 90 ° + θ .$ $B C = h, A B = k .$ $B i s o n e l l i p s e \Rightarrow \underset{- - - - - - - - - - -}{b^{2} p^{2} + a^{2} q^{2} = a^{2} b^{2}}$ $x_{C} = p + h \cos θ; y_{C} = q + h \sin θ$ $x_{A} = p - k \sin θ; y_{A} = q + k \cos θ$ $E l l i p s e e q : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\Rightarrow s l o p e o f a n o r m a l a t (x, y) i s$ $m = - \frac{d x}{d y} = (\frac{a^{2}}{b^{2}}) (\frac{y}{x})$ $m_{B C} = \tan θ = (\frac{a^{2}}{b^{2}}) (\frac{q + h \sin θ}{p + h \cos θ}) . . (i)$ $m_{A B} = - \frac{1}{\tan θ} = (\frac{a^{2}}{b^{2}}) (\frac{q + k \cos θ}{p - k \sin θ}) . . (i i)$ $\Rightarrow (\frac{b^{2} p}{\cos θ} - \frac{a^{2} q}{\sin θ}) = (a^{2} - b^{2}) h . . . (I)$ $\Rightarrow - (\frac{b^{2} p}{\sin θ} + \frac{a^{2} q}{\cos θ}) = (a^{2} - b^{2}) k . . . (I I)$ $N o w C a n d A l i e o n e l l i p s e \Rightarrow$ $\frac{x_{C}^{2}}{a^{2}} + \frac{y_{C}^{2}}{b^{2}} = 1 o r b^{2} {(x_{C})}^{2} + a^{2} {(y_{C})}^{2} = a^{2} b^{2}$ $\Rightarrow b^{2} {(p + h \cos θ)}^{2} + a^{2} {(q + h \sin θ)}^{2} = a^{2} b^{2}$ $a n d a s b^{2} p^{2} + a^{2} q^{2} = a^{2} b^{2} s o w e h a v e$ $2 h (b^{2} p \cos θ + a^{2} q \sin θ)$ $+ h^{2} (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ) = 0$ $\Rightarrow - (\frac{b^{2} p}{\sin θ} + \frac{a^{2} q}{\sin θ}) = \frac{h (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ)}{2 \sin θ \cos θ}$ $N o w u s i n g (I I),$ $(a^{2} - b^{2}) k = \frac{h (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ)}{2 \sin θ \cos θ} . . . (i i i)$ $\frac{x_{A}^{2}}{a^{2}} + \frac{y_{A}^{2}}{b^{2}} = 1 \Rightarrow b^{2} {(x_{A})}^{2} + a^{2} {(y_{A})}^{2} = a^{2} b^{2}$ $\Rightarrow$ $b^{2} {(p - k \sin θ)}^{2} + a^{2} {(q + k \cos θ)}^{2} = a^{2} b^{2}$ $a n d a s b^{2} p^{2} + a^{2} q^{2} = a^{2} b^{2} s o w e h a v e$ $2 k (- b^{2} p \sin θ + a^{2} q \cos θ)$ $+ k^{2} (b^{2} \sin^{2} θ + a^{2} \cos^{2} θ) = 0$ $\Rightarrow \frac{b^{2} p}{\cos θ} - \frac{a^{2} q}{\sin θ} = \frac{k (b^{2} \sin^{2} θ + a^{2} \cos^{2} θ)}{2 \sin θ \cos θ}$ $N o w u s i n g (I),$ $(a^{2} - b^{2}) h = \frac{k (b^{2} \sin^{2} θ + a^{2} \cos^{2} θ)}{2 \sin θ \cos θ} . . . (i v)$ $a n d a l r e a d y w e h a v e$ $(a^{2} - b^{2}) k = \frac{h (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ)}{2 \sin θ \cos θ} . . . (i i i)$ $N o w (i i i) \times (i v) w i t h m = \tan θ g i v e s$ $4 {(a^{2} - b^{2})}^{2} m^{2} = (b^{2} m^{2} + a^{2}) (a^{2} m^{2} + b^{3})$ $\Rightarrow m^{4} + {\frac{a^{4} + b^{4} - 4 {(a^{2} - b^{2})}^{2}}{a^{2} b^{2}}} m^{2} + 1 = 0$ $l e t a / b = μ$ $\Rightarrow m^{4} + {μ^{2} + \frac{1}{μ^{2}} - 4 {(μ - \frac{1}{μ})}^{2}} m^{2} + 1 = 0$ ${m^{2} + 1 - \frac{3}{2} {(μ - \frac{1}{μ})}^{2}}^{2}$ $= {[1 - \frac{3}{2} {(μ - \frac{1}{μ})}^{2}]}^{2} - 1$ $N o w t r a g e d y i s u n l e s s μ = 1, m^{2} \notin R$ $s o m r W S i r k i n d l y h e l p m e$ $o u t o f t h i s t u r m o i l . . . .$ $I f m = \tan θ, w e r e r e a l h a n d k c o u l d$ $e a s i l y b e o b t a i n e d . . .$
	Commented by ajfour last updated on 16/Jul/20

	$t h a n k s s i r, s o m e c o n s o l a t i o n a t$ $l e a s t; b u t t h e r e m u s t b e, o r y o u$ ${c o u l d n}^{'} t h a v e g r a p h e d y o u r s o l u t i o n . .$
	Commented by mr W last updated on 16/Jul/20

	$p e r f e c t s o l u t i o n!$
	Commented by mr W last updated on 17/Jul/20
	$i think all correct! ⇒ m^4 +{μ^2 +(1/μ^2 )−4(μ−(1/μ))^2 }m^2 +1=0 ⇒ m^4 +[2−3(μ−(1/μ))^2 ]m^2 +1=0 ⇒two roots for m^2 for μ≥(√3). for μ=(√3) there should be one single root θ=45°, i.e. m=1. 1^4 +[2−3((√3)−(1/(√3)))^2 ]1^2 +1=0 ⇒true!$
	$i t h i n k a l l c o r r e c t!$ $\Rightarrow m^{4} + {μ^{2} + \frac{1}{μ^{2}} - 4 {(μ - \frac{1}{μ})}^{2}} m^{2} + 1 = 0$ $\Rightarrow m^{4} + [2 - 3 {(μ - \frac{1}{μ})}^{2}] m^{2} + 1 = 0$ $\Rightarrow t w o r o o t s f o r m^{2} f o r μ ⩾ \sqrt{3} .$ $f o r μ = \sqrt{3} t h e r e s h o u l d b e o n e s i n g l e$ $r o o t θ = 45 °, i . e . m = 1 .$ $1^{4} + [2 - 3 {(\sqrt{3} - \frac{1}{\sqrt{3}})}^{2}] 1^{2} + 1 = 0 \Rightarrow t r u e!$
	Commented by ajfour last updated on 17/Jul/20

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