Prove that line y=mx+(3/(4 ))m+(1/m) touches the parabola y^2 =4x+3 whatever the value of m


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Question Number 51236 by peter frank last updated on 25/Dec/18

$P r o v e t h a t l i n e y = m x + \frac{3}{4} m + \frac{1}{m}$ $t o u c h e s t h e p a r a b o l a$ $y^{2} = 4 x + 3 w h a t e v e r t h e$ $v a l u e o f m$
	Answered by mr W last updated on 25/Dec/18

	$y = m x + \frac{3}{4} m + \frac{1}{m} \Rightarrow x = \frac{y}{m} - \frac{3}{4} - \frac{1}{m^{2}}$ $y^{2} = 4 x + 3 \Rightarrow y^{2} = \frac{4 y}{m} - 3 - \frac{4}{m^{2}} + 3$ $\Rightarrow m^{2} y^{2} - 4 m y + 4 = 0$ $\Rightarrow {(m y - 2)}^{2} = 0$ $\Rightarrow y = \frac{2}{m}$ $\Rightarrow x = \frac{1}{m^{2}} - \frac{3}{4}$ $i . e . t h e l i n e a n d t h e p a r a b o l a i n t e r s e c t$ $a l w a y s a t o n e a n d o n l y o n e p o i n t$ $(\frac{1}{m^{2}} - \frac{3}{4}, \frac{2}{m}) f o r a n y v a l u e o f m e x c e p t$ $z e r o . w h e n t w o c u r v e s i n t e r s e c t a t o n e$ $a n d o n l y o n e p o i n t, t h a t m e a n s t h e y$ $t o u c h e a c h o t h e r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18

	$y = m x + \frac{3 m}{4} + \frac{1}{m}$ $y^{2} = 4 x + 3$ $s o l v e . . . w e h a v e t o p r o v e r o o t s a r e s a m e$ $x_{1} = x_{2} a n d y_{1} = y_{2} t h a t m e a n s D = B^{2} - 4 A C = 0$ ${(m x + \frac{3 m}{4} + \frac{1}{m})}^{2} = 4 x + 3$ $m^{2} x^{2} + \frac{9 m^{2}}{16} + \frac{1}{m^{2}} + \frac{6 m^{2} x}{4} + \frac{3}{2} + 2 x = 4 x + 3$ $x^{2} (m^{2}) + x (\frac{6 m^{2}}{4} + 2 - 4) + (\frac{9 m^{2}}{16} + \frac{1}{m^{2}} + \frac{3}{2} - 3) = 0$ $B^{2} = {(\frac{3 m^{2}}{2} - 2)}^{2} = \frac{9 m^{4}}{4} - 2 \times \frac{3 m^{2}}{2} \times 2 + 4 = \frac{9 m^{4}}{4} - 6 m^{2} + 4$ $4 A C = 4 \times m^{2} \times (\frac{9 m^{2}}{16} + \frac{1}{m^{2}} - \frac{3}{2})$ $= \frac{9 m^{4}}{4} + 4 - 6 m^{2}$ $B^{2} = 4 A C$ $h e n c e p r o v e d . . .$
	Commented by peter frank last updated on 25/Dec/18

	$t h a n k y o u s i r . w h a t i s v a l u e$ $o f m$
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