The-line-y-mx-meets-the-parabola-y-x-a-b-x-tangentially-where-0-lt-a-lt-b-Show-that-m-b-a-2-

Question Number 96244 by Don08q last updated on 31/May/20

The line y = m x meets the parabola

y = (x - a) (b - x) tangentially where

0 < a < b . Show that m = {(\sqrt{b} - \sqrt{a})}^{2}

Commented by bobhans last updated on 31/May/20

y = - x^{2} + b x + a x - a b

y = - x^{2} + (a + b) x - a b

w i t h s l o p e = - 2 x + a + b = m

x = \frac{a + b - m}{2} \Rightarrow - x^{2} + (a + b) x - a b = m x

x^{2} + (m - a - b) x + a b = 0

set a + b = p

\Rightarrow {(\frac{p - m}{2})}^{2} + (m - p) (\frac{p - m}{2}) + a b = 0

- {(p - m)}^{2} = - 4 a b \Rightarrow p - m = \pm 2 \sqrt{a b}

m = p \pm 2 \sqrt{a b} = (a + b \pm 2 \sqrt{a b})

m = {(\sqrt{a} \pm \sqrt{b})}^{2}

Commented by Don08q last updated on 31/May/20

Thank you Sir .

Answered by Smail last updated on 31/May/20

m x = (x - a) (b - x) a t t h e i n t e r s e c t i o n

m x = - x^{2} + (b + a) x - a b

x^{2} - (a + b - m) x + a b = 0

S i n c e t h e l i n e i s t a n g e n t

t o t h e p a r a b o l a : Δ = 0 = {(a + b - m)}^{2} - 4 a b

a + b - m = \underline{+} 2 \sqrt{a b}

a + b \underline{+} 2 \sqrt{a b} = m

m = {(\sqrt{a} \underline{+} \sqrt{b})}^{2}

T h u s, t h e r e a r e t w o p a s s i b l e

l i n e s w h o s e s l o p e s a r e m_{1} = {(\sqrt{a} - \sqrt{b})}^{2}

a n d m_{2} = {(\sqrt{a} + \sqrt{b})}^{2}

Commented by Don08q last updated on 31/May/20

Thank you Sir .

Facebook Tweet Pin