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Question Number 53595 by ajfour last updated on 23/Jan/19

Commented by ajfour last updated on 24/Jan/19

Commented by ajfour last updated on 24/Jan/19

$F i n d x, g i v e n α . (t w o r e c t a n g l e s) .$
Commented by mr W last updated on 24/Jan/19

Commented by ajfour last updated on 24/Jan/19

$∠ E A F i s a r i g h t a n g l e t o o, S i r .$ ${i t}^{'} l l b e t h e s a m e e q u a t i o n;$ $1 - x^{4} = k x, i f k = \tan α$ $i j u s t t r i e d t o d e p i c t i t g e o m e t r i c a l y .$
Commented by mr W last updated on 24/Jan/19

$g r e a t i d e a!$
	Answered by ajfour last updated on 23/Jan/19

	Commented by mr W last updated on 24/Jan/19

	$A (R, h)$ $y = {c x}^{2} w i t h c = 1$ $y^{'} = 2 c x$ $x_{P} = R + R \cos θ = R (1 + \cos θ)$ $y_{P} = h - R \sin θ = {c x}_{P}^{2} = {c R}^{2} {(1 + \cos θ)}^{2}$ $\Rightarrow h = R [\sin θ + c R {(1 + \cos θ)}^{2}]$ $y_{P}^{'} = 2 c R (1 + \cos θ) = \frac{1}{\tan θ}$ $\Rightarrow \tan θ + \sin θ = \frac{1}{2 c R} . . . (i)$ $B (r, k)$ $\Rightarrow k = r [\sin ϕ + c r {(1 + \cos ϕ)}^{2}]$ $\Rightarrow \tan ϕ + \sin ϕ = \frac{1}{2 c r} . . . (i i)$ $h - k = \sqrt{{(R + r)}^{2} - {(R - r)}^{2}} = 2 \sqrt{R r}$ $\Rightarrow R [\sin θ + c R {(1 + \cos θ)}^{2}] - r [\sin ϕ + c r {(1 + \cos ϕ)}^{2}] = 2 \sqrt{R r} . . . (i i i)$ $. . . . . . 3 e q n . w i t h 3 u n k n o w n s r, ϕ, θ$ $f r o m (i) :$ $\tan θ + \sin θ = \frac{1}{2 c R} = \frac{4}{k} \Rightarrow k = 8 c R$ $w i t h t = \tan \frac{θ}{2}$ $\frac{2 t}{1 - t^{2}} + \frac{2 t}{1 + t^{2}} = \frac{4}{k}$ $t^{4} + k t - 1 = 0$ $\Rightarrow t = f (k) = . . . .$ $f r o m (i i) :$ $\Rightarrow c r = \frac{1}{2 (\tan ϕ + \sin ϕ)}$ $f r o m (i i i) :$ $\Rightarrow \frac{k [k + 4 t (1 + t^{2})]}{8 {(1 + t^{2})}^{2}} - \frac{1}{\tan ϕ + \sin ϕ} [\sin ϕ + \frac{{(1 + \cos ϕ)}^{2}}{2 (\tan ϕ + \sin ϕ)}] = \sqrt{\frac{k}{\tan ϕ + \sin ϕ}}$ $\Rightarrow ϕ = . . . . .$
	Commented by ajfour last updated on 01/Feb/19

	$T h a t s w h a t s i r, w e c a n t e v e n g e t$ $c e n t r e o f c i r c l e s, g i v e n t h e r a d i i,$ $b i q u a d r a t i c p r o b l e m . .$
	Commented by mr W last updated on 24/Jan/19

	$t^{4} + k t - 1 = 0$ $s h o u l d b e s o l v a b l e . c a n M J S s i r h e l p ?$
	Commented by MJS last updated on 24/Jan/19
	$no “beautiful” solution t_(1, 2) =α±(√β) ∈R t_(3, 4) =γ±(√δ) ∉R α=−γ β=−γ^2 +(k/(4γ)) δ=−γ^2 −(k/(4γ)) γ=(1/((1152))^(1/6) )(√(((9k^2 +(√(81k^4 +768))))^(1/3) −((−9k^2 +(√(81k^4 +768))))^(1/3) )) ∀k∈R\{0} k=0 ⇒ t=±1$
	$no ‘ ‘ {beautiful}^{″} solution$ $t_{1, 2} = α \pm \sqrt{β} \in R$ $t_{3, 4} = γ \pm \sqrt{δ} \notin R$ $α = - γ$ $β = - γ^{2} + \frac{k}{4 γ}$ $δ = - γ^{2} - \frac{k}{4 γ}$ $γ = \frac{1}{\sqrt[6]{1152}} \sqrt{\sqrt[3]{9 k^{2} + \sqrt{81 k^{4} + 768}} - \sqrt[3]{- 9 k^{2} + \sqrt{81 k^{4} + 768}}}$ $\forall k \in R ∖ {0}$ $k = 0 \Rightarrow t = \pm 1$
	Commented by mr W last updated on 24/Jan/19

	$s i r, {i t}^{'} s a beautiful solution!$
	Commented by ajfour last updated on 24/Jan/19

	$t h a n k s a f t e r a l l S i r!$
	Commented by MJS last updated on 24/Jan/19

	${you}^{'} re welcome$
	Commented by MJS last updated on 24/Jan/19

	$trying something else :$ $parabola : y = {a x}^{2}$ $normal in P = (\begin{matrix} p \\ {a p}^{2} \end{matrix}) : y = - \frac{1}{2 a p} x + \frac{2 a^{2} p^{2} + 1}{2 a}$ $horizontal : y = h$ $normal \cap horizontal = C = (\begin{matrix} x_{C} \\ y_{C} \end{matrix})$ $C = (\begin{matrix} p (2 p^{2} + 1 - 2 h) \\ h \end{matrix})$ $∣ C P ∣= x_{C} \Rightarrow we can calculate the 3^{rd} parameter$ $when 2 are given$ $a = \frac{h + \sqrt{4 h^{2} - 3 p^{2}}}{3 p^{2}}$ $h = p (- a p + \sqrt{4 a^{2} p^{2} + 1})$ $p = \frac{\sqrt{2 a h - 1 + \sqrt{16 a^{2} h^{2} - 4 a h + 1}}}{\sqrt{6} a}$ $on the other hand two circles both touching$ $the y - axis and each others can be calculated$ $c_{1} : {(x - r_{1})}^{2} + {(y - h_{1})}^{2} = r_{1}^{2}$ $c_{2} : {(x - r_{2})}^{2} + {(y - h_{2})}^{2} = r_{2}^{2}$ $c_{1} \cap c_{2} = exactly 1 solution \Rightarrow$ $\Rightarrow 4 r_{1} r_{2} = {(h_{1} - h_{2})}^{2}$ $. . . not yet sure where this will lead to . . .$
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