Question-89415

Question Number 89415 by ajfour last updated on 17/Apr/20

Commented by ajfour last updated on 17/Apr/20

E l l i p s e : \frac{x^{2}}{4} + y^{2} = 1

A n d i f ∠ O P A = 90 °, F i n d e q .

o f s h o w n p a r a b o l a .

Commented by Tony Lin last updated on 17/Apr/20

l e t P (2 c o s θ, s i n θ)

(2 c o s θ, s i n θ) \cdot (2 c o s θ - 2, s i n θ) = 0

3 {c o s}^{2} θ - 4 c o s θ + 1 = 0

(3 c o s θ - 1) (c o s θ - 1) = 0

c o s θ = \frac{1}{3} o r 1 (f a l s e)

\Rightarrow P (\frac{2}{3}, \frac{2 \sqrt{2}}{3})

l e t p a r a b o l a y = a {(x - k)}^{2}

{\begin{cases} 1 = {a k}^{2} \\ \frac{2 \sqrt{2}}{3} = a {(\frac{2}{3} - k)}^{2} \end{cases}

Commented by ajfour last updated on 17/Apr/20

t h i s c a n n o t b e e q . o f t h e r e q u i r e d

p a r a b o l a, p l e a s e s o l v e a g a i n, S i r .

Commented by Tony Lin last updated on 17/Apr/20

\Rightarrow {\begin{cases} a = \frac{9}{4} + \frac{3}{\sqrt{2}} - \frac{3 \sqrt{3}}{\sqrt[4]{2}} \\ k = (2 + \frac{4 \sqrt{2}}{3}) (3 + \sqrt{6 \sqrt{2}}) \end{cases}

\Rightarrow y = (\frac{9}{4} + \frac{3}{\sqrt{2}} - \frac{3 \sqrt{3}}{\sqrt[4]{2}}) {x - (2 + \frac{4 \sqrt{2}}{3}) (3 + \sqrt{6 \sqrt{2}})}^{2}

Answered by mr W last updated on 18/Apr/20

\frac{x^{2}}{4} + y^{2} = 1

P (2 \cos θ, \sin θ)

{(2 \cos θ)}^{2} + \sin^{2} θ + 4 {(1 - \cos θ)}^{2} + \sin^{2} θ = 4

3 \cos^{2} θ - 4 \cos θ + 1 = 0

(3 \cos θ - 1) (\cos θ - 1) = 0

\Rightarrow \cos θ = \frac{1}{3} \Rightarrow \sin θ = \frac{2 \sqrt{2}}{3}

P (\frac{2}{3}, \frac{2 \sqrt{2}}{3})

e q n . o f p a r a b o l a : s a y y = A {(x - B)}^{2}

1 = {A B}^{2} \dots (i)

\frac{2 \sqrt{2}}{3} = A {(\frac{2}{3} - B)}^{2} \dots (i i)

{(\frac{2}{3 B} - 1)}^{2} = \frac{2 \sqrt{2}}{3}

B = \frac{2}{3 \pm \sqrt{6 \sqrt{2}}}

\Rightarrow B = \frac{2}{3 + \sqrt{6 \sqrt{2}}} < \frac{2}{3}

A = \frac{1}{B^{2}}

e q n . :

y = \frac{1}{B^{2}} {(x - B)}^{2} = {(\frac{x}{B} - 1)}^{2}

\Rightarrow y = {[\frac{(3 + \sqrt{6 \sqrt{2}}) x}{2} - 1]}^{2}

Commented by mr W last updated on 18/Apr/20

Commented by ajfour last updated on 19/Apr/20

T h a n k s, S i r .

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