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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
$let P(2cosθ,sinθ) (2cosθ,sinθ)∙(2cosθ−2,sinθ)=0 3cos^2 θ−4cosθ+1=0 (3cosθ−1)(cosθ−1)=0 cosθ=(1/3) or 1(false) ⇒P((2/3), ((2(√2))/3)) let parabola y=a(x−k)^2 { ((1=ak^2 )),((((2(√2))/3)=a((2/3)−k)^2 )) :}$
$l e t P (2 c o s θ, s i n θ)$ $(2 c o s θ, s i n θ) \cdot (2 c o s θ - 2, s i n θ) = 0$ $3 {c o s}^{2} θ - 4 c o s θ + 1 = 0$ $(3 c o s θ - 1) (c o s θ - 1) = 0$ $c o s θ = \frac{1}{3} o r 1 (f a l s e)$ $\Rightarrow P (\frac{2}{3}, \frac{2 \sqrt{2}}{3})$ $l e t p a r a b o l a y = a {(x - k)}^{2}$ ${\begin{cases} 1 = {a k}^{2} \\ \frac{2 \sqrt{2}}{3} = a {(\frac{2}{3} - k)}^{2} \end{cases}$
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
$⇒ { ((a=(9/4)+(3/(√2))−((3(√3))/(2)^(1/4) ))),((k=(2+((4(√2))/3))(3+(√(6(√2)))))) :} ⇒y=((9/4)+(3/(√2))−((3(√3))/(2)^(1/4) )){x−(2+((4(√2))/3))(3+(√(6(√2))))}^2$
$\Rightarrow {\begin{cases} a = \frac{9}{4} + \frac{3}{\sqrt{2}} - \frac{3 \sqrt{3}}{\sqrt[4]{2}} \\ k = (2 + \frac{4 \sqrt{2}}{3}) (3 + \sqrt{6 \sqrt{2}}) \end{cases}$ $\Rightarrow y = (\frac{9}{4} + \frac{3}{\sqrt{2}} - \frac{3 \sqrt{3}}{\sqrt[4]{2}}) {x - (2 + \frac{4 \sqrt{2}}{3}) (3 + \sqrt{6 \sqrt{2}})}^{2}$
	Answered by mr W last updated on 18/Apr/20

	$\frac{x^{2}}{4} + y^{2} = 1$ $P (2 \cos θ, \sin θ)$ ${(2 \cos θ)}^{2} + \sin^{2} θ + 4 {(1 - \cos θ)}^{2} + \sin^{2} θ = 4$ $3 \cos^{2} θ - 4 \cos θ + 1 = 0$ $(3 \cos θ - 1) (\cos θ - 1) = 0$ $\Rightarrow \cos θ = \frac{1}{3} \Rightarrow \sin θ = \frac{2 \sqrt{2}}{3}$ $P (\frac{2}{3}, \frac{2 \sqrt{2}}{3})$ $e q n . o f p a r a b o l a : s a y y = A {(x - B)}^{2}$ $1 = {A B}^{2} . . . (i)$ $\frac{2 \sqrt{2}}{3} = A {(\frac{2}{3} - B)}^{2} . . . (i i)$ ${(\frac{2}{3 B} - 1)}^{2} = \frac{2 \sqrt{2}}{3}$ $B = \frac{2}{3 \pm \sqrt{6 \sqrt{2}}}$ $\Rightarrow B = \frac{2}{3 + \sqrt{6 \sqrt{2}}} < \frac{2}{3}$ $A = \frac{1}{B^{2}}$ $e q n . :$ $y = \frac{1}{B^{2}} {(x - B)}^{2} = {(\frac{x}{B} - 1)}^{2}$ $\Rightarrow y = {[\frac{(3 + \sqrt{6 \sqrt{2}}) x}{2} - 1]}^{2}$
	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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