Question-39470

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 \dots

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 \dots

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 \dots 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 =