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Question Number 81854 by ajfour last updated on 16/Feb/20

Commented by ajfour last updated on 16/Feb/20

$F i n d e q . o f p a r a b o l a u s i n g$ $a, b, c i n t h e f o r m y = q - x^{2} .$
	Answered by mr W last updated on 16/Feb/20

	Commented by mr W last updated on 16/Feb/20

	$a n a t t e m p t . . .$ $φ = \cos^{- 1} \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ $x_{B} = a \cos θ$ $y_{B} = a \sin θ$ $\Rightarrow a \sin θ = q - a^{2} \cos^{2} θ . . . (i)$ $x_{A} = b \cos (θ + φ) = b (\cos θ \cos φ - \sin θ \sin φ)$ $y_{A} = b \sin (θ + φ) = b (\sin θ \cos φ + \cos θ \sin φ)$ $\Rightarrow b (\sin θ \cos φ + \cos θ \sin φ) = q - b^{2} {(\cos θ \cos φ - \sin θ \sin φ)}^{2}$ $(i i) - (i) :$ $b (\sin θ \cos φ + \cos θ \sin φ) - a \sin θ$ $= a^{2} \cos^{2} θ - b^{2} {(\cos θ \cos φ - \sin θ \sin φ)}^{2}$ $w e c a n s o l v e t h i s f o r θ a n d t h e n$ $g e t q f r o m (i) :$ $q = a (\sin θ + a^{2} \cos θ)$
	Commented by ajfour last updated on 16/Feb/20
	$equivalently let further A(bcos ψ,bsin ψ) ⇒ a^2 sin^2 θ−asin θ+(q−a^2 )=0 b^2 sin^2 ψ−bsin ψ+(q−b^2 )=0 sin θ=(1/(2a))+(√((1/(4a^2 ))−(q/a^2 )+1)) sin ψ=(1/(2b))+(√((1/(4b^2 ))−(q/b^2 )+1)) ________________________ And θ+ψ = sin^(−1) {(1/(2a))+(√((1/(4a^2 ))−(q/a^2 )+1))} +sin^(−1) {(1/(2b))+(√((1/(4b^2 ))−(q/b^2 )+1))} =π−cos^(−1) (((a^2 +b^2 −c^2 )/(2ab))) _________________________■$
	$e q u i v a l e n t l y$ $l e t f u r t h e r A (b \cos ψ, b \sin ψ)$ $\Rightarrow a^{2} \sin^{2} θ - a \sin θ + (q - a^{2}) = 0$ $b^{2} \sin^{2} ψ - b \sin ψ + (q - b^{2}) = 0$ $\sin θ = \frac{1}{2 a} + \sqrt{\frac{1}{4 a^{2}} - \frac{q}{a^{2}} + 1}$ $\sin ψ = \frac{1}{2 b} + \sqrt{\frac{1}{4 b^{2}} - \frac{q}{b^{2}} + 1}$ $________________________$ $A n d θ + ψ =$ $\sin^{- 1} {\frac{1}{2 a} + \sqrt{\frac{1}{4 a^{2}} - \frac{q}{a^{2}} + 1}}$ $+ \sin^{- 1} {\frac{1}{2 b} + \sqrt{\frac{1}{4 b^{2}} - \frac{q}{b^{2}} + 1}}$ $= π - \cos^{- 1} (\frac{a^{2} + b^{2} - c^{2}}{2 a b})$ $_________________________{◼}$
	Commented by mr W last updated on 16/Feb/20

	$v e r y n i c e w i t h a f i n a l e q n . f o r q!$
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