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 𝛂=tan^(−1) (3/4) .


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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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