Question-81854

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 \dots

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 \dots (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

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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