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Question Number 52558 by ajfour last updated on 09/Jan/19

Commented by ajfour last updated on 10/Jan/19

$e q . o f p a r a b o l a b e y = \frac{x^{2}}{a} a n d$ $e q . o f l i n e b e y = m x (m = \tan ϕ),$ $i f b o t h c i r c l e s t o u c h t h e l i n e a t a$ $s a m e p o i n t a n d c i r c l e i n r e d h a s$ $m a x i m u m r a d i u s, d e t e r m i n e$ $\tan ϕ a n d r a d i u s r .$
Commented by ajfour last updated on 10/Jan/19

Commented by ajfour last updated on 10/Jan/19

$y = \frac{x^{2}}{a} \Rightarrow \tan ϕ = \frac{d y}{d x} = \frac{2 x}{a} . . . (i), (i i)$ $\sin ϕ = \frac{x - r}{3 r} . . . . (i i i)$ $\tan α = \tan (\frac{π}{4} - \frac{ϕ}{2}) = \frac{r}{y + 3 r \cos ϕ} . . (i v)$ $___________________________$ $\sin ϕ = \frac{a \tan ϕ - 2 r}{6 r} [u s i n g (i i) i n (i i i)]$ $. . . . . (I)$ $u s i n g (i) i n (i v) w e g e t$ $\frac{1 - \tan \frac{ϕ}{2}}{1 + \tan \frac{ϕ}{2}} = \frac{r}{\frac{a^{2} \tan^{2} ϕ}{4 a} + 3 r \cos ϕ} . . (I I)$ $l e t \tan \frac{ϕ}{2} = t i n (I)$ $\frac{2 t (6 r)}{1 + t^{2}} = \frac{2 a t}{1 + t^{2}} - 2 r$ $\Rightarrow r (6 t + 1 + t^{2}) = a t$ $\Rightarrow r = \frac{a t}{t^{2} + 6 t + 1} . . . . (I I I)$ $N o w c o n s i d e r i n g (I I)$ $\frac{1 - t}{1 + t} = \frac{r}{\frac{{a t}^{2}}{{(1 + t^{2})}^{2}} + 3 r (\frac{1 - t^{2}}{1 + t^{2}})}$ $\Rightarrow (1 - t) [{a t}^{2} + 3 r (1 - t^{4})] = r (1 + t) {(1 + t^{2})}^{2}$ $\Rightarrow$ ${a t}^{2} (1 - t) = r [(1 + t) {(1 + t^{2})}^{2} - 3 (1 - t) (1 - t^{4})]$ $u s i n g (I I I) i n e q . a b o v e$ $t (1 - t) (t^{2} + 6 t + 1) = (1 + t) {(1 + t^{2})}^{2}$ $- 3 (1 - t) (1 - t^{4})$ $F r o m a b o v e e q . w e o b t a i n$ $t = \tan \frac{ϕ}{2} \approx 0.517$ $\tan ϕ = \frac{2 t}{1 + t^{2}} \approx 0.816$ $& r = \frac{a t}{t^{2} + 6 t + 1} \approx 0.1183 a .$
Commented by MJS last updated on 12/Jan/19

$I^{'} m confused . . .$ ${it}^{'} s possible to get the equations of 2 circles$ $with radius r, having exactly 1 common point$ $and their common tangent t : y = x \tan ϕ$ ${it}^{'} s possible to find a parabola having exactly$ $1 common point with the red circle (at least$ $approximately)$ $but in general, the common tangent of$ $the circle and the parabola is not parallel$ $to y = x \tan ϕ$
Commented by ajfour last updated on 12/Jan/19

$w h e n t h e r e d c i r c l e h a s t o b e o f$ $m a x i m u m r a d i u s, i t s h o u l d b e s o$ $S i r .$
Commented by MJS last updated on 12/Jan/19

${it}^{'} s just not obvious to me . . . I^{'} ll have to go$ $through it . . .$
	Answered by MJS last updated on 12/Jan/19

	$line l : y = k x$ $circle 1 c_{1} : {(x - r)}^{2} + {(y - n)}^{2} = r^{2}$ $circle 2 c_{2} : {(x - p)}^{2} + {(y - q)}^{2} = r^{2}$ $O = l \cap c_{1} \cap c_{2} = (\begin{matrix} (1 + \frac{k}{\sqrt{k^{2} + 1}}) r \\ k (1 + \frac{k}{\sqrt{k^{2} + 1}}) r \end{matrix})$ $n = (k + \sqrt{k^{2} + 1}) r$ $p = (1 + \frac{2 k}{\sqrt{k^{2} + 1}}) r$ $q = (k + \frac{k^{2} - 1}{\sqrt{k^{2} + 1}}) r$ $parabola p a r : y = \frac{1}{a} x^{2}$ $P = p a r \cap c_{2}$ $(1) c_{2} : {(x - p)}^{2} + {(y - q)}^{2} = r^{2}$ $(2) p a r : y = \frac{1}{a} x^{2}$ $(3) derivate of (1)$ $(4) derivate of (2)$ $variables a, k, r, x, y$ $(5) maximum for r$ $not sure if this can be solved$ $it looks to me as if for given a we can find$ $matching k, x, y for any value of r \Rightarrow no$ $maximum exists$ $can you exactly draw your solution to$ $make it clearer ?$
	Commented by MJS last updated on 12/Jan/19

	$imagine the following steps$ $1 . draw y = \frac{1}{a} x^{2}$ $2 . fix a pole at origin, attach 2 circles with$ $any radius$ $3 . slide up the circles along the pole, the$ $left handed circle along the y - axis$ $\Rightarrow at some distance the right handed circle$ $will intersect the parabola in exactly 1 point$ $why should there be an upper limit for r ?$
	Commented by ajfour last updated on 13/Jan/19

	$y o u h a v e a G e n u i n e o b j e c t i o n S i r .$
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