Math Question and Answers


Question and Answers Forum

All Questions Topic List
Coordinate Geometry Questions
Previous in All Question Next in All Question
Previous in Coordinate Geometry Next in Coordinate Geometry
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

$it seems not unique to me at first sight but$ ${let}^{'} s try . . .$
Commented by ajfour last updated on 14/Oct/18
$let eq. of ellipse be ax^2 +by^2 +2hxy+2gx+2fy+c=0 since it has a double root with y=0 ⇒ the equation ax^2 +2gx+c= 0 has D=0 ⇒ g^2 = ac ....(i) double root with x=0 ⇒ by^2 +2fy+c = 0 has D=0 ⇒ f^( 2) = bc ...(ii) Let slope of PQ= tan θ = m { ((x_Q = 2cos^2 θ )),((y_Q = 2cos θsin θ)) :} .....(iii) if (p,q) be center of ellipse ⇒ q = pm ⇒ ((af−gh)/(h^2 −ab)) = m(((bg−hf)/(h^2 −ab))) ...(iv) OQ=r = (√(p^2 +q^2 ))−2cos θ { ((x_Q = p−rcos θ)),((y_Q = q−rsin θ)) :} ......(v) differentiating eq. of ellipse 2ax+2by(dy/dx)+2hx(dy/dx)+2hy+2g+ 2f (dy/dx) = 0 (dy/dx)∣_Q = −(1/m) ....(vi) And Q(x_Q ,y_Q ) lies on ellipse so (x_Q , y_Q ) satisfies ax^2 +by^2 +2hxy+2gx+2fy+c=0 ....(vii) ________________________$
$l e t e q . o f e l l i p s e b e$ ${a x}^{2} + {b y}^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ $s i n c e i t h a s a d o u b l e r o o t w i t h$ $y = 0 \Rightarrow t h e e q u a t i o n$ ${a x}^{2} + 2 g x + c = 0 h a s D = 0$ $\Rightarrow g^{2} = a c . . . . (i)$ $d o u b l e r o o t w i t h x = 0 \Rightarrow$ ${b y}^{2} + 2 f y + c = 0 h a s D = 0$ $\Rightarrow f^{2} = b c . . . (i i)$ $L e t s l o p e o f P Q = \tan θ = m$ ${\begin{cases} x_{Q} = 2 \cos^{2} θ \\ y_{Q} = 2 \cos θ \sin θ \end{cases} . . . . . (i i i)$ $i f (p, q) b e c e n t e r o f e l l i p s e$ $\Rightarrow q = p m$ $\Rightarrow \frac{a f - g h}{h^{2} - a b} = m (\frac{b g - h f}{h^{2} - a b}) . . . (i v)$ $O Q = r = \sqrt{p^{2} + q^{2}} - 2 \cos θ$ ${\begin{cases} x_{Q} = p - r \cos θ \\ y_{Q} = q - r \sin θ \end{cases} . . . . . . (v)$ $d i f f e r e n t i a t i n g e q . o f e l l i p s e$ $2 a x + 2 b y \frac{d y}{d x} + 2 h x \frac{d y}{d x} + 2 h y + 2 g +$ $2 f \frac{d y}{d x} = 0$ $\frac{d y}{d x} ∣_{Q} = - \frac{1}{m} . . . . (v i)$ $A n d Q (x_{Q}, y_{Q}) l i e s o n e l l i p s e$ $s o (x_{Q}, y_{Q}) s a t i s f i e s$ ${a x}^{2} + {b y}^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ $. . . . (v i i)$ $________________________$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum