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Question Number 47135 by Tawa1 last updated on 05/Nov/18

Commented by Meritguide1234 last updated on 05/Nov/18

$a n s {2, 1} a n d {\frac{2}{5}, - \frac{1}{5}}$
Commented by behi83417@gmail.com last updated on 06/Nov/18
$x^2 +y^2 =(1/t)⇒ { ((5x(1+t)=12)),((5y(1−t)=4)) :} ((25)/t)=((144)/((1+t)^2 ))+((16)/((1−t)^2 ))⇒ 25(1−t^2 )^2 =t[16+32t+16t^2 +144−288t+144t^2 ] 25−50t^2 +25t^4 =160t−256t^2 +160t^3 25t^4 −160t^3 +206t^2 −160t+25=0 ⇒t^4 −6.4t^3 +8.24t^2 −6.4t+1=0 ⇒t=5,0.2 t=5⇒ { ((x=((12)/(5(1+5)))=(2/5))),((y=(4/(5(1−5)))=−(1/5))) :} t=0.2⇒ { ((x=((12)/(5(1+0.2)))=2)),((y=(4/(5(1−0.2)))=1.■)) :}$
$x^{2} + y^{2} = \frac{1}{t} \Rightarrow {\begin{cases} 5 x (1 + t) = 12 \\ 5 y (1 - t) = 4 \end{cases}$ $\frac{25}{t} = \frac{144}{{(1 + t)}^{2}} + \frac{16}{{(1 - t)}^{2}} \Rightarrow$ $25 {(1 - t^{2})}^{2} = t [16 + 32 t + 16 t^{2} + 144 - 288 t + 144 t^{2}]$ $25 - 50 t^{2} + 25 t^{4} = 160 t - 256 t^{2} + 160 t^{3}$ $25 t^{4} - 160 t^{3} + 206 t^{2} - 160 t + 25 = 0$ $\Rightarrow t^{4} - 6.4 t^{3} + 8.24 t^{2} - 6.4 t + 1 = 0$ $\Rightarrow t = 5, 0.2$ $t = 5 \Rightarrow {\begin{cases} x = \frac{12}{5 (1 + 5)} = \frac{2}{5} \\ y = \frac{4}{5 (1 - 5)} = - \frac{1}{5} \end{cases}$ $t = 0.2 \Rightarrow {\begin{cases} x = \frac{12}{5 (1 + 0.2)} = 2 \\ y = \frac{4}{5 (1 - 0.2)} = 1 . ◼ \end{cases}$
Commented by Tawa1 last updated on 06/Nov/18

$God bless you sir$
	Answered by behi83417@gmail.com last updated on 06/Nov/18
	$x=rcost,y=rsint ⇒ { ((5rcost(1+(1/r^2 ))=12)),((5rsint(1−(1/r^2 ))=4)) :} 25r^2 =((144)/((1+(1/r^2 ))^2 ))+((16)/((1−(1/r^2 ))^2 ))⇒ 25r^2 (r^4 −1)^2 =144r^4 (r^2 −1)^2 +16r^4 (r^2 +1)^2 25(r^8 −2r^4 +1)=144r^2 (r^4 −2r^2 +1)+16r^2 (r^4 +2r^2 +1) 25r^8 −50r^4 +25=144r^6 −288r^4 +144r^2 +16r^6 +32r^4 +16r^2 25r^8 −160r^6 +206r^4 −160r^2 +25=0 r^8 −6.4r^6 +8.24r^4 −6.4r^2 +1=0 r=±2.24,±0.45 r=2.24⇒cost=((12(2.24)^2 )/(5×(±2.24)(2.24^2 +1)))=±0.89 ⇒t=27.13^• ,152.87^• ⇒ { ((x=2.24×(±0.89)#±2.0)),((y=2.24×0.46#1.0)) :} r=0.45⇒cost=((12(.45)^2 )/(5×(±.45)(.45^2 +1)))=±0.9 t=25.84^• ,154.16^• ⇒ { ((x=0.45×(±0.9)=±0.4)),((y=0.45×0.45=0.2)) :} ⇒(x,y)=(±2,1),(±0.4,0.2) .■$
	$x = r c o s t, y = r s i n t$ $\Rightarrow {\begin{cases} 5 r c o s t (1 + \frac{1}{r^{2}}) = 12 \\ 5 r s i n t (1 - \frac{1}{r^{2}}) = 4 \end{cases}$ $25 r^{2} = \frac{144}{{(1 + \frac{1}{r^{2}})}^{2}} + \frac{16}{{(1 - \frac{1}{r^{2}})}^{2}} \Rightarrow$ $25 r^{2} {(r^{4} - 1)}^{2} = 144 r^{4} {(r^{2} - 1)}^{2} + 16 r^{4} {(r^{2} + 1)}^{2}$ $25 (r^{8} - 2 r^{4} + 1) = 144 r^{2} (r^{4} - 2 r^{2} + 1) + 16 r^{2} (r^{4} + 2 r^{2} + 1)$ $25 r^{8} - 50 r^{4} + 25 = 144 r^{6} - 288 r^{4} + 144 r^{2} + 16 r^{6} + 32 r^{4} + 16 r^{2}$ $25 r^{8} - 160 r^{6} + 206 r^{4} - 160 r^{2} + 25 = 0$ $r^{8} - 6.4 r^{6} + 8.24 r^{4} - 6.4 r^{2} + 1 = 0$ $r = \pm 2.24, \pm 0.45$ $r = 2.24 \Rightarrow c o s t = \frac{12 {(2.24)}^{2}}{5 \times (\pm 2.24) (2 . 24^{2} + 1)} = \pm 0.89$ $You can't use 'macro parameter character #' in math mode$ $r = 0.45 \Rightarrow c o s t = \frac{12 {(. 45)}^{2}}{5 \times (\pm . 45) (. 45^{2} + 1)} = \pm 0.9$ $t = 25 . 84^{∙}, 154 . 16^{∙} \Rightarrow {\begin{cases} x = 0.45 \times (\pm 0.9) = \pm 0.4 \\ y = 0.45 \times 0.45 = 0.2 \end{cases}$ $\Rightarrow (x, y) = (\pm 2, 1), (\pm 0.4, 0.2) . ◼$
	Commented by Tawa1 last updated on 05/Nov/18

	$God bless you sir$
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