Two lines through the point (1,−3) are tamgent to the curve y=x^2 . Find the equation of these two lines and make a sketch to verify your results.


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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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