Q.63108 (A check) eq. of ellipse (x^2 /4)+y^2 =1 Inscribed equilateral △ABC of side s=((16(√6))/(√(365))) Do these points satisfy for A, B, C ? A((4/(√(365))), ((19)/(√(365)))) ; B[−(((20+8(√3)))/(√(365))), ((8(√3)−5)/(√(365)))] C[−(((20−8(√3)))/(√(365))) , −(((5+8(√3)))/(√(365)))] θ=45°


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Question Number 63218 by ajfour last updated on 30/Jun/19

$Q . 63108 (A c h e c k)$ $e q . o f e l l i p s e$ $\frac{x^{2}}{4} + y^{2} = 1$ $I n s c r i b e d e q u i l a t e r a l △ A B C$ $o f s i d e s = \frac{16 \sqrt{6}}{\sqrt{365}}$ $D o t h e s e p o i n t s s a t i s f y f o r$ $A, B, C ?$ $A (\frac{4}{\sqrt{365}}, \frac{19}{\sqrt{365}});$ $B [- \frac{(20 + 8 \sqrt{3})}{\sqrt{365}}, \frac{8 \sqrt{3} - 5}{\sqrt{365}}]$ $C [- \frac{(20 - 8 \sqrt{3})}{\sqrt{365}}, - \frac{(5 + 8 \sqrt{3})}{\sqrt{365}}]$ $θ = 45 °$
Commented by Tinkutara@ last updated on 02/Jul/19
hello Mr W ! remember me? btw are you active on WhatsApp, Facebook, Snapchat, Twitter, LinkedIn, Pinterest ? p.s- recently, my Vacations got over. �� ( went to phi phi Island!!! { Thailand})
Commented by ajfour last updated on 01/Jul/19
$A,B,C satisfy the ellipse eq., and AB=BC=CA=s ; I checked it sir, please agree.. cos^2 θ=(1/2)+(1/((a^2 −b^2 ))){((q/2)+(√((q^2 /4)−(p^3 /(27)))))^(1/3) +((q/2)−(√((q^2 /4)−(p^3 /(27)))))^(1/3) } p=(3/(16))(a^2 −b^2 )^2 q=(((a^2 +b^2 )^3 )/(32))+(3/8)a^2 b^2 (a^2 +b^2 )−((3a^4 b^4 )/s^2 ) .$
$A, B, C s a t i s f y t h e e l l i p s e e q .,$ $a n d A B = B C = C A = s;$ $I c h e c k e d i t s i r, p l e a s e a g r e e . .$ $\cos^{2} θ = \frac{1}{2} + \frac{1}{(a^{2} - b^{2})} {{(\frac{q}{2} + \sqrt{\frac{q^{2}}{4} - \frac{p^{3}}{27}})}^{1 / 3} + {(\frac{q}{2} - \sqrt{\frac{q^{2}}{4} - \frac{p^{3}}{27}})}^{1 / 3}}$ $p = \frac{3}{16} {(a^{2} - b^{2})}^{2}$ $q = \frac{{(a^{2} + b^{2})}^{3}}{32} + \frac{3}{8} a^{2} b^{2} (a^{2} + b^{2}) - \frac{3 a^{4} b^{4}}{s^{2}} .$
Commented by kaivan.ahmadi last updated on 01/Jul/19

$h i m r w w h e r e a r e y o u f r o m ?$ $d o y o u h a v e i n s t a g r a m p a g e ?$
Commented by mr W last updated on 01/Jul/19

$y e s s i r! a f t e r n e w c o n s i d e r a t i o n n o w$ $i t h i n k s o l u t i o n s o t h e r t h a n$ $θ = 0 ° a r e p o s s i b l e . i^{'} l l t r y t o f i n d o u t$ $t h e m i s t a k e s i n m y w o r k i n g .$
Commented by Tinkutara@ last updated on 02/Jul/19
thanks sir! I wish the same for you...
Commented by ajfour last updated on 04/Jul/19
Tinkutara, how did you perform in jee advanced, where are you probably going to join.
Commented by Tinkutara@ last updated on 03/Jul/19
my paper went okayish. Unfortunately , I 'm not getting CSE in top 5 IITS ( due to reservation , sigh!). So Most probably I will be going to Bits Pilani CSE . ( since it's better than rest iits ) . - At the end of the day , I'm thankful to you and rest of the members of this awesome community to provide elegant solutions to the silly doubts thrown at you guys :P
Commented by Tinkutara last updated on 18/Jul/19
@ajfour Sir: I got a rank between 5900-6k in advanced and got IIT KGP. I am also thankful to all of them who solved my doubts on this platform which meant a lot to me! But @above is not my id.
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