Question-90520

Question Number 90520 by ajfour last updated on 24/Apr/20

Commented by ajfour last updated on 24/Apr/20

I f e q . o f e l l i p s e i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,

f i n d e q u a t i o n o f s h o w n p a r a b o l a .

Answered by mr W last updated on 25/Apr/20

l e t μ = \frac{b}{a}

l e t e q n . o f p a r a b o l a b e

y = A {(x - h)}^{2} + B

b = A {(- h)}^{2} + B

0 = A {(a - h)}^{2} + B

\Rightarrow \frac{B}{A} = - {(a - h)}^{2}

\Rightarrow A = \frac{μ}{2 h - a}

\Rightarrow B = - \frac{μ {(a - h)}^{2}}{2 h - a}

b^{2} x^{2} + a^{2} {[A {(x - h)}^{2} + B]}^{2} - a^{2} b^{2} = 0

b^{2} x^{2} + a^{2} A^{2} x^{4} + 4 h^{2} a^{2} A^{2} x^{2} + a^{2} A^{2} h^{4} - 4 {h a}^{2} A^{2} x^{3} + 2 h^{2} a^{2} A^{2} x^{2} - 4 h^{3} a^{2} A^{2} x + 2 a^{2} {A B x}^{2} - 2 h 2 a^{2} A B x + 2 a^{2} {A B h}^{2} + a^{2} B^{2} - a^{2} b^{2} = 0

A^{2} x^{4} - 4 {h A}^{2} x^{3} + (6 h^{2} A^{2} + 2 A B + μ^{2}) x^{2} - 4 h A (h^{2} A + B) x + (A^{2} h^{4} + 2 {A B h}^{2} + B^{2} - b^{2}) = 0

x = 0 i s a r o o t :

\Rightarrow A^{2} h^{4} + 2 {A B h}^{2} + B^{2} - b^{2} = 0

\Rightarrow A^{2} x^{3} - 4 {h A}^{2} x^{2} + (6 h^{2} A^{2} + 2 A B + μ^{2}) x - 4 h A (h^{2} A + B) = 0

x = a, p, p a r e r o o t s :

x^{3} - 4 {h x}^{2} + (6 h^{2} + 2 \frac{B}{A} + \frac{μ^{2}}{A}) x - 4 h (h^{2} + \frac{B}{A})

= (x - a) {(x - p)}^{2}

= x^{3} - (a + 2 p) x^{2} + (2 a + p) p x - {a p}^{2}

(a + 2 p) = 4 h \dots (i)

(2 a + p) p = (6 h^{2} + 2 \frac{B}{A} + \frac{μ^{2}}{A}) \dots (i i)

{a p}^{2} = 4 h (h^{2} + \frac{B}{A}) \dots (i i i)

f r o m (i) a n d (i i i) w e g e t

\Rightarrow 16 h^{2} - 8 a h - a^{2} = 0

\Rightarrow h = \frac{(1 + \sqrt{2}) a}{4}

Commented by mr W last updated on 25/Apr/20

Commented by mr W last updated on 25/Apr/20

Commented by mr W last updated on 25/Apr/20

Commented by ajfour last updated on 25/Apr/20

B e a u t i f u l, S i r . T h a n k s a l o t!

Answered by ajfour last updated on 25/Apr/20

l e t e q . o f p a r a b o l a b e

y = c (x - a) (x - \frac{b}{a c})

e l l i p s e : b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}

f o r t h e i r c o m m o n p o i n t s,

w h i c h a r e x = 0, x = a, x = p, x = p

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

\Rightarrow a^{2} c^{2} x^{4} - 2 a^{2} c^{2} (a + \frac{b}{a c}) x^{3}

+ a^{2} c^{2} [\frac{2 b}{c} + \frac{b^{2}}{a^{2} c^{2}} + {(a + \frac{b}{a c})}^{2}] x^{2}

- 2 a^{2} b c (a + \frac{b}{a c}) x = 0

N o w a + 2 p = 2 (a + \frac{b}{a c})

a n d {a p}^{2} = \frac{2 b}{c} (a + \frac{b}{a c})

\Rightarrow a {(a + \frac{2 b}{a c})}^{2} = \frac{8 b}{c} (a + \frac{b}{a c})

\Rightarrow {(a^{2} c + 2 b)}^{2} = 8 b (a^{2} c + b)

o r a^{4} c^{2} - 4 a^{2} b c - 4 b^{2} = 0

\Rightarrow c = \frac{2 b}{a^{2}} (1 + \sqrt{2}), h e n c e

e q . o f p a r a b o l a i s

y = \frac{2 b}{a^{2}} (1 + \sqrt{2}) (x - a) [x - \frac{b}{a} \cdot \frac{a^{2}}{2 b (1 + \sqrt{2})}]

o r

y = 2 (1 + \sqrt{2}) (\frac{b}{a^{2}}) (x - a) {x - \frac{a (\sqrt{2} - 1)}{2}}

Commented by ajfour last updated on 25/Apr/20

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