Question-45477

Question Number 45477 by ajfour last updated on 13/Oct/18

Commented by ajfour last updated on 13/Oct/18

I f r a d i u s o f c i r c l e i s u n i t y,

f i n d e q u a t i o n o f e l l i p s e .

(c i r c l e m a y o r m a y n o t t o u c h y - a x e s)

Commented by ajfour last updated on 13/Oct/18

Commented by MJS last updated on 13/Oct/18

it seems not unique to me at first sight but

{let}^{'} s try \dots

Commented by ajfour last updated on 14/Oct/18

l e t e q . o f e l l i p s e b e

{a x}^{2} + {b y}^{2} + 2 h x y + 2 g x + 2 f y + c = 0

s i n c e i t h a s a d o u b l e r o o t w i t h

y = 0 \Rightarrow t h e e q u a t i o n

{a x}^{2} + 2 g x + c = 0 h a s D = 0

\Rightarrow g^{2} = a c \dots . (i)

d o u b l e r o o t w i t h x = 0 \Rightarrow

{b y}^{2} + 2 f y + c = 0 h a s D = 0

\Rightarrow f^{2} = b c \dots (i i)

L e t s l o p e o f P Q = \tan θ = m

{\begin{cases} x_{Q} = 2 \cos^{2} θ \\ y_{Q} = 2 \cos θ \sin θ \end{cases} \dots . . (i i i)

i f (p, q) b e c e n t e r o f e l l i p s e

\Rightarrow q = p m

\Rightarrow \frac{a f - g h}{h^{2} - a b} = m (\frac{b g - h f}{h^{2} - a b}) \dots (i v)

O Q = r = \sqrt{p^{2} + q^{2}} - 2 \cos θ

{\begin{cases} x_{Q} = p - r \cos θ \\ y_{Q} = q - r \sin θ \end{cases} \dots \dots (v)

d i f f e r e n t i a t i n g e q . o f e l l i p s e

2 a x + 2 b y \frac{d y}{d x} + 2 h x \frac{d y}{d x} + 2 h y + 2 g +

2 f \frac{d y}{d x} = 0

\frac{d y}{d x} ∣_{Q} = - \frac{1}{m} \dots . (v i)

A n d Q (x_{Q}, y_{Q}) l i e s o n e l l i p s e

s o (x_{Q}, y_{Q}) s a t i s f i e s

{a x}^{2} + {b y}^{2} + 2 h x y + 2 g x + 2 f y + c = 0

\dots . (v i i)

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