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Question Number 42624 by ajfour last updated on 29/Aug/18

Commented by ajfour last updated on 29/Aug/18

$T h e s m a l l e r c i r c l e h a s r a d i u s r .$ $T h e t w o p a r a b o l a s a r e r e f l e c t i o n$ $o f o n e a n o t h e r i n x - a x i s .$ $E q u a t i o n o f u p p e r p a r a b o l a i s$ $y = {a x}^{2} + r . F i n d r a d i u s R o f$ $l a r g e r c i r c l e i n t e r m s o f a, r .$
Commented by MJS last updated on 02/Sep/18

${p a r a b o l a}^{+} = P^{+} : y = {a x}^{2} + r$ $r e d {c i r c l e}^{+} = C^{+} : y = \sqrt{R^{2} - {(x - (R + r))}^{2}}$ $1 . attempt$ $P^{+} \cap C^{+} must have exactly one real solution$ $2 . attempt$ $tangent of P^{+} = tangent of C^{+}$ $3 . attempt$ $normal of P^{+} = normal of C^{+} must intersect$ $x - axis at x = R + r$ $all of these lead to polynomes of 3^{rd} or 4^{th} degree$ $which we cannot solve generally (too many$ $parameters)$ $so I tried something different :$ $given are the circles$ $C_{1} : x^{2} + y^{2} = 1$ $C_{2} : {(x - (r + 1))}^{2} + y^{2} = r^{2}$ $find the parabola P : y = {a x}^{2} + 1 touching C_{2}$ $in exactly one point \Leftrightarrow P and C_{2} have a common$ $tangent in T = (\begin{matrix} t \\ {a t}^{2} + 1 \end{matrix}) = (\begin{matrix} t \\ \sqrt{r^{2} - {(t - (r + 1))}^{2}} \end{matrix})$ $this also leads to a polynome of 4^{th} degree$ $which can be generally solved; the ‘ ‘ {usual}^{″}$ $complicated solution using$ $(t - α - \sqrt{β}) (t - α + \sqrt{β}) (t - γ - \sqrt{δ} i) (t - γ + \sqrt{δ} i) = 0$ $with only one of the real pair α \pm \sqrt{β} solving$ $the original unsquared equation$ $we can approximately solve for t with any$ $given r and then easily get a .$ $r > 1 \Leftrightarrow a > 0$ $r = 1 \Leftrightarrow a = 0 (P degenerates into the line y = 1)$ $r < 1 \Leftrightarrow a < 0$ $interestingly this has got at least one solution$ $with r, t, a \in Q : r = \frac{5}{27} t = \frac{4}{3} a = - \frac{1}{2}$ $will post later$
	Answered by ajfour last updated on 03/Sep/18
	$eq. of parabola y = ax^2 +r P(x_1 , y_1 ) eq. of larger circle (R+r−x)^2 +y^2 = R^2 distance of P from centre of larger circle is R ⇒ R^2 =(R+r−x)^2 +(ax^2 +r)^2 2R((dR/dx))=2(R+r−x)((dR/dx)−1)+ 2(ax^2 +r)(2ax) As R is minimum, (dR/dx) must be zero: ⇒ R+r−x = 2ax(ax^2 +r) ⇒ x = f(R) ⇒ R^2 =[R+r−f(R)]^2 +{a[f(R)]^2 +r}^2 R is then obtained from above eq. ________________________$
	$e q . o f p a r a b o l a y = {a x}^{2} + r$ $P (x_{1}, y_{1})$ $e q . o f l a r g e r c i r c l e$ ${(R + r - x)}^{2} + y^{2} = R^{2}$ $d i s t a n c e o f P f r o m c e n t r e o f$ $l a r g e r c i r c l e i s R$ $\Rightarrow R^{2} = {(R + r - x)}^{2} + {({a x}^{2} + r)}^{2}$ $2 R (\frac{d R}{d x}) = 2 (R + r - x) (\frac{d R}{d x} - 1) +$ $2 ({a x}^{2} + r) (2 a x)$ $A s R i s m i n i m u m, \frac{d R}{d x} m u s t b e z e r o :$ $\Rightarrow R + r - x = 2 a x ({a x}^{2} + r)$ $\Rightarrow x = f (R)$ $\Rightarrow R^{2} = {[R + r - f (R)]}^{2} + {a {[f (R)]}^{2} + r}^{2}$ $R i s t h e n o b t a i n e d f r o m a b o v e e q .$ $________________________$
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