Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 39470 by Raj Singh last updated on 06/Jul/18

Commented by maxmathsup by imad last updated on 07/Jul/18

$x^{2} + y^{2} = 1 \Rightarrow y^{2} = 1 - x^{2} \Rightarrow 1 - x^{2} = 4 x - 4 \Rightarrow x^{2} + 4 x - 5 = 0 \Rightarrow$ $Δ^{^{'}} = 2^{2} + 5 = 9 \Rightarrow x_{1} = - 2 + 3 = 1 a n d x_{2} = - 2 - 3 = - 5$ $x = 1 \Rightarrow y = 0 \Rightarrow A (1, 0) i s i n t e r s e c t i o n$ $x = - 5 \Rightarrow y^{2} = - 24 a n d t h i s i s i m p o s s i b l e s o t h e f i r s t p o i n t i s A .$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

	Answered by ajfour last updated on 06/Jul/18

	$x^{2} + 4 (x - 1) = 1$ $x^{2} + 4 x - 5 = 0$ ${(x + 2)}^{2} = 9$ $x = - 2 \pm 3; b u t f o r p a r a b o l a x ⩾ 1$ $h e n c e o n l y p o i n t i s (1, 0); h e n c e$ $t h e y t o u c h (c i r c l e b e i n g a c l o s e d$ $c u r v e) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

	$s o l v e . . .$ $x^{2} + 4 x - 4 = 1$ $x^{2} + 5 x - x - 5 = 0$ $x (x + 5) - 1 (x + 5) = 0$ $(x + 5) (x - 1) = 0$ $x = 1, - 5$ $w h e n x = 1 y = 0$ $25 + y^{2} = 1 t h e n y^{2} = - 24 w h i c h i s n o t p o s s i b l e .$ $a c i r c l e a n d p a r a b o l a i n t e s e c t a t t w o d i s t i n c t$ $p o i n t s B u t h e r e o n l y o n e s o l u t i o n (1, 0)$ $s a t i s f y b o t h e q u a t i o n . . .$ $s o t h i z t w o c u r v e c i r c l e a n d p a r a b o l a c a n n o t$ $i n t e r s e c t a t s i n g l e p o i n t . . . H e n c e t h e y t o u c h$ $e a c h o t h e r a t (1, 0)$
	Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

	$w h e n t o u c h e a c h o t h e r s l o p e v a l u e c a c u l a t e d s t t h a t w i t h$ $r e s p e c t t o b o t h c u r v e s h o u l d b e s a m e$ $x^{2} + y^{2} = 1$ $2 x + 2 y . \frac{d y}{d x} = 0$ $\frac{d y}{d x} a t (1, 0) i s m_{1} = {(\frac{- x}{y})}_{1^{'} 0} = \infty$ $y^{2} = 4 (x - 1)$ $2 y \frac{d y}{d x} = 4 \frac{d y}{d x} = \frac{2}{y} s o s l o p e a t (1, 0) i s = \infty =$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum