Prove-that-line-y-mx-3-4-m-1-m-touches-the-parabola-y-2-4x-3-whatever-the-value-of-m-

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 \dots 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 \dots

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