Question-63336

Question Number 63336 by ajfour last updated on 02/Jul/19

Commented by ajfour last updated on 02/Jul/19

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

i n t h e f o r m : y = {A x}^{2} + C .

Answered by mr W last updated on 07/Jul/19

y = {A x}^{2} + C

y^{'} = 2 A x

P [a (1 + \cos θ), a (1 - \sin θ)]

\tan θ = \frac{1}{2 A a (1 + \cos θ)}

l e t t = \tan \frac{θ}{2}

\frac{2 t}{1 - t^{2}} = \frac{1}{2 A a (1 + \frac{1 - t^{2}}{1 + t^{2}})}

\frac{2 t}{1 - t^{2}} = \frac{1 + t^{2}}{4 A a}

\Rightarrow t^{4} + 8 A a t - 1 = 0

\Rightarrow 8 A = \frac{1 - t^{4}}{a t}

a (1 - \sin θ) = {A a}^{2} {(1 + \cos θ)}^{2} + C

a \frac{{(1 - t)}^{2}}{1 + t^{2}} = 4 {A a}^{2} \frac{1}{{(1 + t^{2})}^{2}} + C \dots (i)

Q [b (1 + \cos φ), - b (1 + \sin φ)]

\tan φ = \frac{1}{2 A b (1 + \cos φ)}

l e t s = \tan \frac{φ}{2}

\Rightarrow s^{4} + 8 A b s - 1 = 0

\Rightarrow 8 A = \frac{1 - s^{4}}{b s}

- b (1 + \sin φ) = {A b}^{2} {(1 + \cos φ)}^{2} + C

- b \frac{{(1 + s)}^{2}}{1 + s^{2}} = 4 {A b}^{2} \frac{1}{{(1 + s^{2})}^{2}} + C \dots (i i)

(i) - (i i) :

a \frac{{(1 - t)}^{2}}{1 + t^{2}} + b \frac{{(1 + s)}^{2}}{1 + s^{2}} = \frac{1 - t^{4}}{2 a t} [\frac{a^{2}}{{(1 + t^{2})}^{2}} - \frac{b^{2}}{{(1 + s^{2})}^{2}}] \dots (i i i)

\frac{1 - t^{4}}{a t} = \frac{1 - s^{4}}{b s} \dots (i v)

\dots \dots

e x a m p l e :

a = 5, b = 4

\Rightarrow t = \tan \frac{θ}{2} = 0.10001

\Rightarrow s = \tan \frac{φ}{2} = 0.12489

Commented by mr W last updated on 07/Jul/19

Commented by mr W last updated on 08/Jul/19

i c a n n o t f i n d a w a y t o r e d u c e t h e

s o l u t i o n t o a s i n g l e e q u a t i o n .

Commented by ajfour last updated on 08/Jul/19

I t s a l r i g h t s i r, b u t l e t s i n q u i r e

i n s o m e o t h e r f a s h i o n t o o, w e

m i g h t h i t l u c k . .

Answered by ajfour last updated on 08/Jul/19

Commented by ajfour last updated on 08/Jul/19

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