find the complex form of equation 4x^2 −2y^2 =5


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Question Number 113262 by bemath last updated on 12/Sep/20

$find the complex form of$ $equation {4 x}^{2} - {2 y}^{2} = 5$
	Answered by mr W last updated on 12/Sep/20

	$\frac{x^{2}}{{(\frac{\sqrt{5}}{2})}^{2}} - \frac{y^{2}}{{(\frac{\sqrt{5}}{\sqrt{2}})}^{2}} = 1$ $c^{2} = a^{2} + b^{2} = \frac{5}{4} + \frac{5}{2} = \frac{15}{4}$ $c = \frac{\sqrt{15}}{2}$ ${i t}^{'} s a h y p e r b o l a w i t h f o c i (- \frac{\sqrt{15}}{2}, 0)$ $a n d (\frac{\sqrt{15}}{2}, 0) a s w e l l a s s e m i m a j o r$ $a x i s a = \frac{\sqrt{5}}{2} .$ $a c c o r d i n g t o t h e d e f i n i t i o n o f$ $h y p e r b o l a w e c a n w r i t e i t s e q u a t i o n$ $a l s o i n c o m p l e x f o r m :$ $∣ z + c ∣ - ∣ z - c ∣= \pm 2 a$ $o r$ $∣ z + \frac{\sqrt{15}}{2} ∣ - ∣ z - \frac{\sqrt{15}}{2} ∣= \pm \sqrt{5}$
	Commented by mr W last updated on 12/Sep/20

	$c e r t a i n l y y o u c a n a l s o w r i t e t h e$ $e q u a t i o n i n m a n y o t h e r c o m p l e x f o r m s .$
	Commented by bemath last updated on 12/Sep/20

	$thank you sir$
	Answered by john santu last updated on 12/Sep/20
	$let z = x+iy and z^− = x−iy { ((z+z^− = (2x,0); x = ((z+z^− )/2))),((z−z^− = (0, 2y); y=((z−z^− )/2))) :} ⇔ 4x^2 −2y^2 =5 in complex form 4(((z+z^− )/2))^2 −2(((z−z^− )/2))^2 = 5 (z+z^− )^2 −(((z−z^− )^2 )/2) = 5 2(z+z^− )^2 −(z−z^− )^2 = 10$
	$l e t z = x + i y a n d \bar{z} = x - i y$ ${\begin{cases} z + \bar{z} = (2 x, 0); x = \frac{z + \bar{z}}{2} \\ z - \bar{z} = (0, 2 y); y = \frac{z - \bar{z}}{2} \end{cases}$ $\Leftrightarrow 4 x^{2} - 2 y^{2} = 5 i n c o m p l e x f o r m$ $4 {(\frac{z + \bar{z}}{2})}^{2} - 2 {(\frac{z - \bar{z}}{2})}^{2} = 5$ ${(z + \bar{z})}^{2} - \frac{{(z - \bar{z})}^{2}}{2} = 5$ $2 {(z + \bar{z})}^{2} - {(z - \bar{z})}^{2} = 10$
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