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Question Number 213888 by mnjuly1970 last updated on 20/Nov/24

Commented by mnjuly1970 last updated on 20/Nov/24

$c i r c l e i s t a n g a n t t o p a r a b o l a$ $. A i s c e n t e r o f c i r c l e$ $F i n d, i n f (R) = ?$
Commented by mr W last updated on 21/Nov/24

$y = x^{r} ?$ $A (0, α) ?$ $i n f (R) m e a n s w h a t ?$
Commented by mehdee7396 last updated on 21/Nov/24

$y = x^{2}$ $A (0, a)$
Commented by mnjuly1970 last updated on 21/Nov/24

$i n f i m u m$ $i n f {x ∣ x \in [0, 1)} = i n f {x ∣ x \in (0, 1)} = 1$
Commented by mr W last updated on 21/Nov/24

$i {d o n}^{'} t u n d e r s t a n d t h e s e n s e t o a s k$ $f o r i n f (R) h e r e .$
	Answered by mr W last updated on 21/Nov/24

	$y = x^{2}$ $x^{2} + {(y - a)}^{2} = R^{2}$ $y + {(y - a)}^{2} = R^{2}$ $y^{2} - (2 a - 1) y + a^{2} - R^{2} = 0$ $Δ = {(2 a - 1)}^{2} - 4 (a^{2} - R^{2}) = 0$ $4 R^{2} = 4 a - 1$ $\Rightarrow R = \sqrt{a - \frac{1}{4}}$ $a ⩾ R = \sqrt{a - \frac{1}{4}}$ $a^{2} - a + \frac{1}{4} ⩾ 0$ $\Rightarrow a ⩾ \frac{1}{2}$
	Commented by mr W last updated on 21/Nov/24

	Commented by mnjuly1970 last updated on 21/Nov/24

	$t h a n k s a l o t : i n f (R) = 1 / 2$
	Commented by mr W last updated on 21/Nov/24

	Answered by BHOOPENDRA last updated on 21/Nov/24

	$E q u a t i o n o f c i r c l e$ $w h i c h c e n t e r A (0, a)$ $x^{2} + {(y - a)}^{2} = R^{2}$ $t a n g e n c y$ $y = x^{2} S o$ $x^{2} + {(x^{2} - a)}^{2} = R^{2}$ $p u t x^{2} = y$ $F o r t h e c i r c l e t o b e t a n g e n t t o t h e p a r a b o l a$ $t h e q u d r a t i c m u s t h a v e a r e p e a t e d$ $r o o t . T h e d i s c r i m i n a t o f q u d r a t i c$ $m u s t b e z e r o .$ $Δ = b^{2} - 4 a c$ ${(2 a - 1)}^{2} = 4 (a^{2} - R^{2})$ $R^{2} = (a - \frac{1}{4})$ $T h e q u e s t i o n a s k f o r t h e s m a l l e s t$ $p o s s i b l e v a l u e o f t h e r a d i u s R o f$ $a c i r c l e c e n t e r e d a t A (0, a) t h a t$ $t o u c h e s t h e p a r a b o l a y = x^{2} a t e x a c t l y$ $t w o p o i n t s .$ $T h e i n f i m u m o f R i s s m a l l e s t v a l u e R$ $c a n a p p r o a c h (b u t n o t n e c e s s r i l y r e a c h) .$
	Commented by mnjuly1970 last updated on 21/Nov/24

	$\underset{―}{}$
	Commented by BHOOPENDRA last updated on 21/Nov/24

	$K e y o b s e r v a t i o n$ $f o r R > 0, t h e c o n d i t i o n i s : a > \frac{1}{4}$ $T h e r a d i u s R m i n i m i z e d w h e n a$ $a p p r o a c h e s i t s l o w e r b o u n d, a = \frac{1}{4} .$ $A t t h i s v a l u e$ $R^{2} = 1 / 4 - 1 / 4 = 0$ $W h i l e t h e i n f i m u m o f R^{2} m a t h e m a t i c a l l y 0$ $, t h e s m a l l e s t p h y s i c a l r a d i u s R i s$ $a c h i e v e d j u s t a b o v e a = \frac{1}{4}, w h e r e$ $R s t a r t i n c r e a s i n g f r o m 0 .$ $I n f i m u m o f t h e r a d i u s R i s 0$ $B u t i t w i l l n o t s a t i s f y t a n g e n t i a l$ $c o n d i t i o n$ $S o R^{2} = a - 1 / 4 p u t i n t h e e q u a t i o n$ $y^{2} + (2 a - 1) y + (a^{2} - R^{2}) = 0$ $y^{2} + (2 a - 1) y + (a^{2} - a + \frac{1}{4}) = 0$ $y^{2} + (2 a - 1) y + {(a - \frac{1}{2})}^{2} = 0$ $a - \frac{1}{2} ⩾ 0$ $a ⩾ \frac{1}{2}$ $R ⩾ \frac{1}{2}$ $I n f (R) = 1 / 2 (a c c o r d i n g t o t a n g e n t i a l$ $c o n d i t i o n)$
	Commented by mr W last updated on 21/Nov/24

	$a c i r c l e w i t h r a d i u s 0 i s n o t a c i r c l e,$ $b u t j u s t a p o i n t . s u c h t h a t t h e c i r c l e$ $t a n g e n t s t h e p a r a b o l l a a s r e q u e s t e d,$ $a ⩾ \frac{1}{2} a n d c o r r e s p o n d i n g l y R ⩾ \frac{1}{2} .$
	Commented by BHOOPENDRA last updated on 21/Nov/24

	$I d i d n o t s a y t h i s S i r W i w a s j u s t e x p l a n i n g$ $i n f (R) . i a l r e d y s a i d t h i s R c a n n e v e r$ $b e z e r o$
	Commented by mr W last updated on 21/Nov/24

	$i n c u r r e n t c a s e ‘ ‘ a > \frac{1}{4} a n d R > 0^{″}$ $i s n o t t r u e, i t s h o u l d b e ‘ ‘ a ⩾ \frac{1}{2} a n d$ $R ⩾ {\frac{1}{2}}^{″} .$
	Commented by BHOOPENDRA last updated on 21/Nov/24

	$w h a t e v e r y o u g o t i t s c o r r e c t b u t$ $I n t h e c o n t e x t o f i n t e r s e c t i o n p o i n t$ $y^{2} - (2 a - 1) y + (a^{2} - R^{2}) = 0$ $s u b s t i t u t e R^{2} = a - 1 / 4$ $y^{2} - (2 a - 1) y + (a^{2} - a + \frac{1}{4}) = 0$ $y^{2} - (2 a - 1) y + {(a - \frac{1}{2})}^{2} = 0$ $f o r t h i s q u d r a t i c t o h a v e r e a l s o l u t i o n$ $a - \frac{1}{2},$ $\Rightarrow a ⩾ \frac{1}{2}$ $R ⩾ \frac{1}{2}$ $i k n o w t h i s$ $i w a s j u s t e x p l a n i n g i n f (R) .$ $B T W t h a n k s$
	Commented by mr W last updated on 21/Nov/24
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