How-do-you-find-the-1-Focal-point-2-Directrix-for-y-ax-2-bx-c-For-simplicity-lets-assume-it-goes-throigh-point-0-0-

Question Number 5113 by FilupSmith last updated on 14/Apr/16

How do you find the :

(1) Focal point

(2) Directrix

for y = {a x}^{2} + b x + c

For simplicity, lets assume it goes throigh

point (0, 0) .

Answered by Yozzii last updated on 14/Apr/16

F o r a p a r a b o l a o f t h e f o r m 4 p y = x^{2},

i t s f o c u s i s F (0, p) a n d i t s d i r e c t r i x i s

g i v e n b y y = - p .

W e c a n r e w r i t e y = {a x}^{2} + b x + c a s

y = a {(x + \frac{b}{2 a})}^{2} + c - \frac{b^{2}}{4 a}

o r \frac{1}{a} (y + \frac{b^{2}}{4 a} - c) = {(x + \frac{b}{2 a})}^{2} .

L e t J = y + \frac{b^{2}}{4 a} - c, N = x + \frac{b}{2 a} .

\frac{1}{a} J = N^{2} . J a n d N r e p r e s e n t t h e

c o o r d i n a t e s y s t e m (N, J) .

\Rightarrow 4 p = \frac{1}{a} \Rightarrow p = \frac{1}{4 a} . S o, i n (N, J) c o o r d i n a t e s

F (0, \frac{1}{4 a}) a n d d i r e c t r i x J = - \frac{1}{4 a} . S i n c e N = x + \frac{b}{2 a}

a n d J = y + \frac{b^{2}}{4 a} - c w e c a n c o n v e r t b a c k

t o x y - c o o r d i n a t e s g i v i n g

F (- \frac{b}{2 a}, \frac{1 - b^{2}}{4 a} + c) a n d d i r e c t r i x y = c - \frac{1 + b^{2}}{4 a} .

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