Find-the-equation-of-ellipse-with-F-1-1-2-F-2-3-4-and-2a-2-3-

Question Number 136165 by bramlexs22 last updated on 19/Mar/21

F i n d t h e e q u a t i o n o f e l l i p s e w i t h

F_{1} (1, 2), F_{2} (3, 4) a n d 2 a = 2 \sqrt{3}

Commented by bramlexs22 last updated on 19/Mar/21

D e a r M r W c a n y o u h e l p m e

Commented by mr W last updated on 19/Mar/21

d o y o u m e a n t h e f o c i w i t h F_{1} a n d F_{2} ?

Commented by bramlexs22 last updated on 19/Mar/21

y e s s i r

Answered by mr W last updated on 19/Mar/21

p e r d e f i n i t i o n t h e s u m o f d i s t a n c e s

f r o m a n y p o i n t o n a n e l l i p s e t o t h e

f o c i i s c o n s t a n t w h i c h i s 2 a .

\sqrt{{(x - 1)}^{2} + {(y - 2)}^{2}} + \sqrt{{(x - 3)}^{2} + {(y - 4)}^{2}} = 2 \sqrt{3}

\sqrt{{(x - 1)}^{2} + {(y - 2)}^{2}} = 2 \sqrt{3} - \sqrt{{(x - 3)}^{2} + {(y - 4)}^{2}}

{(x - 1)}^{2} + {(y - 2)}^{2} = 12 + {(x - 3)}^{2} + {(y - 4)}^{2} - 4 \sqrt{3} \sqrt{{(x - 3)}^{2} + {(y - 4)}^{2}}

x + y - 8 = \sqrt{3} \sqrt{{(x - 3)}^{2} + {(y - 4)}^{2}}

x^{2} + y^{2} + 64 + 2 x y - 16 x - 16 y = 3 (x^{2} - 6 x + 9) + 3 (y^{2} - 8 y + 16)

\Rightarrow 2 x^{2} + 2 y^{2} - 2 x y - 2 x - 8 y + 11 = 0

Commented by bramlexs22 last updated on 19/Mar/21

i t r y b y r o t a t i o n w i t h a n g l e

45 ° . i^{'} m s t u c k . f o r g o t b y d e f i n i t i o n . t h a n k s s i r

Commented by mr W last updated on 19/Mar/21

Commented by mr W last updated on 19/Mar/21

i n t h i s c a s e y o u c a n a l s o e a s i l y g e t

t h e e q u a t i o n t h r o u g h r o t a t i o n,

b e c a u s e t h e r o t a t i o n a n g l e i s 45 ° .

Commented by bramlexs22 last updated on 19/Mar/21

y e s s i r b u t i^{'} m s t u c k . i {d o n}^{'} t

w h i c h m y m i s t a k e

Commented by bramlexs22 last updated on 19/Mar/21

can sir show me my mistake

Commented by bramlexs22 last updated on 19/Mar/21

(\begin{matrix} x \\ y \end{matrix}) = {(\begin{matrix} \cos 45 ° - \sin 45 ° \\ \sin 45 ° \cos 45 ° \end{matrix})}^{- 1} (\begin{matrix} x^{'} \\ y^{'} \end{matrix})

Commented by mr W last updated on 19/Mar/21

\boldsymbol M e t h o d \boldsymbol u s i n g \boldsymbol r o t a t i o n

F_{1} F_{2} = \sqrt{{(3 - 1)}^{2} + {(4 - 2)}^{2}} = 2 \sqrt{2}

m i d p o i n t f r o m F_{1} a n d F_{2} i s (2, 3)

a = \sqrt{3} (g i v e n)

b = \sqrt{a^{2} - {(\frac{F_{1} F_{2}}{2})}^{2}} = \sqrt{3 - 2} = 1

e q n . b e f o r e r o t a t i o n :

\frac{{(x - 2)}^{2}}{3} + {(y - 3)}^{2} = 1

r o t a t i o n b y 45 ° ↶ a b o u t p o i n t (2, 3)

x_{1} = (x - 2) \frac{\sqrt{2}}{2} + (y - 3) \frac{\sqrt{2}}{2} + 2 = (x + y - 5) \frac{\sqrt{2}}{2} + 2

y_{1} = - (x - 2) \frac{\sqrt{2}}{2} + (y - 3) \frac{\sqrt{2}}{2} + 3 = (- x + y - 1) \frac{\sqrt{2}}{2} + 3

e q n . a f t e r r o t a t i o n :

\frac{{(x + y - 5)}^{2}}{6} + \frac{{(- x + y - 1)}^{2}}{2} = 1

o r

{(x + y - 5)}^{2} + 3 {(- x + y - 1)}^{2} = 6

x^{2} + y^{2} + 25 + 2 x y - 10 x - 10 y + 3 (x^{2} + y^{2} + 1 - 2 x y + 2 x - 2 y) = 6

\Rightarrow 2 x^{2} + 2 y^{2} - 2 x y - 2 x - 8 y + 11 = 0

Commented by mr W last updated on 19/Mar/21

Commented by bramlexs22 last updated on 19/Mar/21

o o m y m i s t a k e u s e r o t a t i t o n

a b o u t p o i n t (0, 0) s i r . g r e a t s i r