Question-43331

Question Number 43331 by ajfour last updated on 09/Sep/18

Commented by ajfour last updated on 09/Sep/18

F i n d e q u a t i o n o f r e d p a r a b o l a,

i f i t b e r e f l e c t i o n o f y = x^{2} i n

t h e l i n e y = m x + c .

Commented by ajfour last updated on 09/Sep/18

Commented by ajfour last updated on 09/Sep/18

p = \frac{c + m h - k}{\sqrt{1 + m^{2}}}

y - k = 2 p \cos θ

h - x = 2 p \sin θ

\tan \boldsymbol θ = \boldsymbol m

\Rightarrow \boldsymbol x = \boldsymbol h - \frac{2 \boldsymbol m (\boldsymbol c + \boldsymbol m h - \boldsymbol k)}{1 + \boldsymbol m^{2}}

a n d \boldsymbol y = \boldsymbol k + \frac{2 (\boldsymbol c + \boldsymbol m h - \boldsymbol k)}{1 + \boldsymbol m^{2}}

L e t s c o n s i d e r a p o i n t P (h, k) o f

r e d p a r a b o l a .

\Rightarrow f (h, k) = 0

l e t i m a g e o f P (h, k) b e I (x, y)

B u t y = x^{2}, \Rightarrow

\boldsymbol k + \frac{2 (\boldsymbol c + \boldsymbol m h - \boldsymbol k)}{1 + \boldsymbol m^{2}} = {[\boldsymbol h - \frac{2 \boldsymbol m (\boldsymbol c + \boldsymbol m h - \boldsymbol k)}{1 + \boldsymbol m^{2}}]}^{2}

h e n c e f (x, y) i s

y + \frac{2 (c + m x - y)}{1 + m^{2}} = {[x - \frac{2 m (c + m x - y)}{1 + m^{2}}]}^{2} .

J u s t t o c h e c k

l e t m = 1, c = 0

\Rightarrow x = y^{2} (r e f l e c t i o n o f y = x^{2}) .

Commented by ajfour last updated on 10/Sep/18

T h a n k y o u S i r .

Commented by MrW3 last updated on 09/Sep/18

c o r r e c t s i r .

G e n e r a l l y m i r r o r i m a g e o f p o i n t P (x, y) i n

l i n e \boldsymbol a x + \boldsymbol b y + \boldsymbol c = 0 i s p o i n t Q (u, v) w i t h

u = x - \frac{2 a (a x + b y + c)}{a^{2} + b^{2}}

v = y - \frac{2 b (a x + b y + c)}{a^{2} + b^{2}}

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