what-equation-of-ellips-with-F-1-1-2-F-2-3-4-and-a-3-

Question Number 78177 by jagoll last updated on 15/Jan/20

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

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

Commented by MJS last updated on 15/Jan/20

what are we “ allowed to {do}^{″} ?

Commented by jagoll last updated on 15/Jan/20

y e s s i r i {c a n}^{'} t s o l v e t h i s p r o b l e m

Commented by jagoll last updated on 15/Jan/20

w h a t i s c e n t e r p o i n t a t (\frac{1 + 3}{2}, \frac{2 + 4}{2}) = (2, 3) s i r ?

Commented by jagoll last updated on 15/Jan/20

Commented by MJS last updated on 15/Jan/20

one possibility is to take the definition of

an ellipse as “ {recipe}^{″}

ell = {X \in R^{2} ∣ \overset{―}{{X F}_{1}} + \overset{―}{{X F}_{2}} = 2 a}

let F_{1} = (\begin{matrix} p_{1} \\ q_{1} \end{matrix}); F_{2} = (\begin{matrix} p_{2} \\ q_{2} \end{matrix})

\Rightarrow

\sqrt{{(x - p_{1})}^{2} + {(y - q_{1})}^{2}} + \sqrt{{(x - p_{2})}^{2} + {(y - q_{2})}^{2}} = 2 a

\sqrt{α} + \sqrt{β} = 2 a ∣^{2} - (α + β)

2 \sqrt{α} \sqrt{β} = 4 a^{2} - α + β ∣^{2}

4 α β = 16 a^{4} - 8 a^{2} (α + β) + {(α + β)}^{2}

16 a^{4} - 8 a^{2} (α + β) + {(α - β)}^{2} = 0

\dots

{A x}^{2} + B x y + {C y}^{2} + D x + E y + F = 0

A = 4 (2 a + p_{1} - p_{2}) (2 a - p_{1} + p_{2})

B = 8 (p_{1} - q_{2}) (p_{2} - q_{1})

C = 4 (2 a + q_{1} - q_{2}) (2 a - q_{1} + q_{2})

D = 4 ((p_{1} - p_{2}) (p_{1}^{2} + q_{1}^{2} - p_{2}^{2} - q_{2}^{2}) - 4 a^{2} (p_{1} + p_{2}))

E = 4 ((q_{1} - q_{2}) ((p_{1}^{2} + q_{1}^{2} - p_{2}^{2} - q_{2}^{2}) - 4 a^{2} (q_{1} + q_{2}))

F = - 16 a^{4} + 8 a^{2} (p_{1}^{2} + q_{1}^{2} + p_{2}^{2} + q_{2}^{2}) - {(p_{1}^{2} + q_{1}^{2} - p_{2}^{2} - q_{2}^{2})}^{2}

in our case

a = \sqrt{3}; p_{1} = 1; q_{1} = 2; p_{2} = 3; q_{2} = 4

A = 32; B = - 32; C = 32; D = - 32; E = - 128; F = 176 p

\Rightarrow

32 x^{2} - 32 x y + 32 y^{2} - 32 x - 128 y + 176 = 0

x^{2} - x y + y^{2} - x - 4 y + \frac{11}{2} = 0

Commented by MJS last updated on 15/Jan/20

\dots you are right with the center point

Commented by MJS last updated on 15/Jan/20

the other possibility is to calculate b

e = \frac{1}{2} \overset{―}{F_{1} F_{2}} = \sqrt{2}

b = \sqrt{a^{2} - e^{2}} = 1

the equation of the ellipse similar to the

given one but with center in (\begin{matrix} 0 \\ 0 \end{matrix}) and axes

parallel to x - and y - axes is

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

x^{2} + 3 y^{2} - 3 = 0

now we have to rotate and shift it

θ = ∡ (\vec{F_{1} F_{2}}) = 45 °

\Rightarrow {\begin{cases} x^{*} = \frac{\sqrt{2}}{2} (x - y) \\ y^{*} = \frac{\sqrt{2}}{2} (x + y) \end{cases} \Leftrightarrow {\begin{cases} x = \frac{\sqrt{2}}{2} (x^{*} + y^{*}) \\ y = \frac{\sqrt{2}}{2} (y^{*} - x^{*}) \end{cases}

inserting into our equation

2 {(x^{*})}^{2} - 2 x^{*} y^{*} + 2 {(y^{*})}^{2} - 3 = 0

shifting the center

{\begin{cases} x^{*} = x - 2 \\ y^{*} = y - 3 \end{cases}

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

x^{2} - x y + y^{2} - x - 4 y + \frac{11}{2} = 0

Commented by jagoll last updated on 15/Jan/20

s i r w h y c e n t e r p o i n t a t (0, 0) ?

a n d i g o t t h e a n g l e α = \tan^{- 1} (2) . w h e r e

i^{'} m w r o n g s i r ?

Commented by MJS last updated on 15/Jan/20

I rotate first, then shift

you shift first, then rotate

this makes no difference

the angle is the same as the angle of the vector

\vec{F_{1} F_{2}} = (\begin{matrix} 2 \\ 2 \end{matrix}); \tan α = \frac{y}{x} = 1

Commented by jagoll last updated on 15/Jan/20

y e s s i r . t h a n k s

Answered by jagoll last updated on 15/Jan/20

2 c = \sqrt{2^{2} + 2^{2}} \Rightarrow c = \sqrt{2}

b^{2} = a^{2} - c^{2} = 3 - 2 = 1

e l l i p s e q u a t i o n o r i g i n a l l y ?

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

t h i s i s e l l i p s e b y r o t a t e d w i t h

a n g l e α = \tan^{- 1} (2) ?

Commented by MJS last updated on 15/Jan/20

angle α = \tan^{- 1} (1) = 45 °

Commented by MJS last updated on 15/Jan/20

\dots you should get the same result as me . you

shift first, then rotate

Commented by jagoll last updated on 15/Jan/20

o o \tan α = \frac{4 - 2}{3 - 1} = 1 . o o i u n d e r s t a n d

s i r . t h a n k s y o u v e r y m u c h

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