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Question Number 31113 by NECx last updated on 02/Mar/18

T w o l i n e s t h r o u g h t h e p o i n t (1, - 3)

a r e t a m g e n t t o t h e c u r v e y = x^{2} .

F i n d t h e e q u a t i o n o f t h e s e t w o

l i n e s a n d m a k e a s k e t c h t o v e r i f y

y o u r r e s u l t s .

Answered by Tinkutara last updated on 02/Mar/18

Commented by Tinkutara last updated on 02/Mar/18

L e t e q u a t i o n o f t a n g e n t i s y = m x + c

I t t o u c h e s y = x^{2} s o D = 0

m x + c = x^{2}

D = 0 g i v e s c = \frac{- m^{2}}{4}

y = m x - \frac{m^{2}}{4}

I t p a s s e s t h r o u g h (1, - 3)

- 3 = m - \frac{m^{2}}{4}

m = - 2, 6

E q u a t i o n s o f l i n e a r e

y = - 2 x - 1

y = 6 x - 9

Commented by NECx last updated on 03/Mar/18

I^{'} m m o s t g r a t e f u l s i r . T h a n k s .

Answered by mrW2 last updated on 02/Mar/18

e q n . o f l i n e t h r o u g h p o i n t (1, - 3) :

y + 3 = m (x - 1)

o r y = m (x - 1) - 3

i n t e r s e c t i o n :

y = m (x - 1) - 3 = x^{2}

\Rightarrow x^{2} - m x + (m + 3) = 0

D = m^{2} - 4 (m + 3) = 0

\Rightarrow {(m - 2)}^{2} - 16 = 0

\Rightarrow m - 2 = \pm 4

\Rightarrow m = 6 o r - 2

E q n . o f l i n e s :

y = 6 (x - 1) - 3 \Rightarrow y = 6 x - 9

y = - 2 (x - 1) - 3 \Rightarrow y = - 2 x - 1

Commented by NECx last updated on 03/Mar/18

T h a n k y o u s o m u c h s i r .

Commented by NECx last updated on 02/Mar/18

a n d w h a t d o e s D r e p r e s e n t s ?

Commented by mrW2 last updated on 02/Mar/18

I f t h e e q n . {a x}^{2} + b x + c = 0 s h o u l d

h a v e o n l y o n e s o l u t i o n, t h e n

D = b^{2} - 4 a c m u s t e q u a l t o z e r o .

Commented by NECx last updated on 02/Mar/18

w h y i s D = 0 ?

Commented by mrW2 last updated on 02/Mar/18

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