What-is-the-transformed-equation-of-a-parabola-given-by-y-2x-2-8-5-x-109-50-if-the-coordinate-axes-is-rotated-anticlockwise-by-tan-1-3-4-

Question Number 14664 by ajfour last updated on 03/Jun/17

W h a t i s t h e t r a n s f o r m e d

e q u a t i o n o f a p a r a b o l a g i v e n b y

y = 2 x^{2} + \frac{8}{5} x - \frac{109}{50}, i f t h e

c o o r d i n a t e a x e s i s r o t a t e d

a n t i c l o c k w i s e b y α = \tan^{- 1} \frac{3}{4} .

Answered by mrW1 last updated on 03/Jun/17

a f t e r s u c h a r o t a t i o n

x^{'} = x \cos α + y \sin α

y^{'} = - x \sin α + y \cos α

o r

x = x^{'} \cos α - y \sin α

y = x^{'} \sin α + y \cos α

w i t h α = \tan^{- 1} \frac{3}{4}

\Rightarrow \tan α = \frac{3}{4}

\Rightarrow \sin α = \frac{3}{5}

\Rightarrow \cos α = \frac{4}{5}

x = \frac{4}{5} x^{'} - \frac{3}{5} y^{'}

y = \frac{3}{5} x^{'} + \frac{4}{5} y^{'}

\Rightarrow \frac{3}{5} x^{'} + \frac{4}{5} y^{'} = 2 {(\frac{4}{5} x^{'} - \frac{3}{5} y^{'})}^{2} + \frac{8}{5} (\frac{4}{5} x^{'} - \frac{3}{5} y^{'}) - \frac{109}{50}

\Rightarrow 10 (3 x^{'} + 4 y^{'}) = 4 {(4 x^{'} - 3 y^{'})}^{2} + 16 (4 x^{'} - 3 y^{'}) - 109

\Rightarrow 30 x^{'} + 40 y^{'} = (64 x^{' 2} - 96 x^{'} y^{'} + 36 y^{' 2}) + (64 x^{'} - 48 y^{'}) - 109

\Rightarrow 64 x^{' 2} - 96 x^{'} y^{'} + 36 y^{' 2} + 34 x^{'} - 88 y^{'} - 109 = 0

Commented by ajfour last updated on 03/Jun/17

y e s s i r, t h a n k s f o r t h e e f f o r t .

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