Math Question and Answers


Question and Answers Forum

All Questions Topic List
Coordinate Geometry Questions
Previous in All Question Next in All Question
Previous in Coordinate Geometry Next in Coordinate Geometry
Question Number 48156 by ajfour last updated on 20/Nov/18

Commented by ajfour last updated on 20/Nov/18

$F i n d R i n t e r m s o f a .$
	Answered by ajfour last updated on 20/Nov/18

	$L e t c e n t e r o f r e d c i r c l e b e (R, k)$ $a n d c e n t e r o f b l u e o n e (h, R)$ $\Rightarrow P (\frac{h + R}{2}, \frac{k + R}{2})$ $P l i e s o n p a r a b o l a$ $\Rightarrow \frac{k + R}{2} = a {(\frac{h + R}{2})}^{2} . . . . (i)$ $y = {a x}^{2} \Rightarrow - \frac{d x}{d y} = \frac{- 1}{2 a x}$ $a t P : \frac{d x}{d y} = \frac{1}{a (h + R)}$ $\Rightarrow \frac{k - R}{h - R} = \frac{1}{a (h + R)} . . . . (i i)$ $A l s o {(h - R)}^{2} + {(k - R)}^{2} = 4 R^{2}$ $l e t h = R + 2 R \cos θ$ $k = R + 2 R \sin θ$ $u s i n g t h e s e i n (i)$ $1 + \sin θ = a R {(1 + \cos θ)}^{2} . . (I)$ $u s i n g t h e s a m e i n (i i)$ $2 a R (1 + \cos θ) \sin θ = \cos θ . . (I I)$ $(I) \times (I I) g i v e s$ $2 \sin θ (1 + \sin θ) = \cos θ (1 + \cos θ)$ $l e t \tan \frac{θ}{2} = t, t h e n$ $(\frac{4 t}{1 + t^{2}}) (1 + \frac{2 t}{1 + t^{2}}) = (\frac{1 - t^{2}}{1 + t^{2}}) (1 + \frac{1 - t^{2}}{1 + t^{2}})$ $\Rightarrow 2 t {(1 + t)}^{2} = 1 - t^{2}$ $\Rightarrow t = - 1 (i d o n t t h i n k a c c e p t a b l e)$ $o r 2 t (1 + t) = 1 - t$ $\Rightarrow 2 t^{2} + 3 t - 1 = 0$ $t = \frac{- 3 \pm \sqrt{9 + 8}}{4}$ $t a k i n g t = \frac{\sqrt{17} - 3}{4}$ $A n d f r o m (I I)$ $2 a R = \frac{1}{\tan θ (1 + \cos θ)}$ $= \frac{1 - t^{2}}{2 t (1 + \frac{1 - t^{2}}{1 + t^{2}})} = \frac{1 - t^{4}}{4 t}$ $2 a R = \frac{1 - {(\frac{1 - 3 t}{2})}^{2}}{4 t}$ $= \frac{4 - 1 - 9 t^{2} + 6 t}{16 t}$ $= \frac{3 + 6 t - 9 (\frac{1 - 3 t}{2})}{16 t}$ $= \frac{6 + 12 t - 9 + 27 t}{32 t}$ $= \frac{39 t - 3}{32 t} = \frac{39}{32} - \frac{3}{32 t}$ $= \frac{39}{32} - \frac{3 \times 4}{32 (\sqrt{17} - 3)}$ $= \frac{39}{32} - \frac{12 (\sqrt{17} + 3)}{32 \times 8}$ $2 a R = \frac{78 - 3 \sqrt{17} - 9}{64}$ $o r R = \frac{3 (23 - \sqrt{17})}{128 a} .$
	Commented by ajfour last updated on 20/Nov/18

	$A n y e r r o r i n t h i s s o l u t i o n S i r ?$
	Commented by mr W last updated on 20/Nov/18

	$a l l c o r r e c t s i r!$
	Answered by mr W last updated on 20/Nov/18

	$P (h, k)$ $k = {a h}^{2}$ $\tan θ = 2 a h$ $e q n . o f t a n g e n t :$ $y = 2 a h (x - h) + {a h}^{2}$ $x = 0 : y = - {a h}^{2}$ $y = 0 : x = h / 2$ $l = \frac{h}{2 \cos θ}$ $φ = π / 2 - θ$ $R = l \tan (θ / 2)$ $R = 2 l \tan (φ / 2) = 2 l \tan (π / 4 - θ / 2)$ $\Rightarrow \tan (θ / 2) = 2 \tan (π / 4 - θ / 2) = \frac{2 (1 - \tan θ / 2)}{1 + \tan θ / 2}$ $w i t h t = \tan (θ / 2)$ $t + t^{2} = 2 - 2 t$ $t^{2} + 3 t - 2 = 0$ $\Rightarrow t = \frac{\sqrt{17} - 3}{2}$ $a l = \frac{a h}{2 \cos θ} = \frac{\tan θ}{4 \cos θ}$ $a R = a l \tan \frac{θ}{2} = \frac{\tan θ \tan \frac{θ}{2}}{4 \cos θ} = \frac{t^{2} (1 + t^{2})}{2 {(1 - t^{2})}^{2}}$
	Commented by ajfour last updated on 20/Nov/18

	$I t s o k a y S i r, t h a n k y o u .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum