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 66161 by peter frank last updated on 09/Aug/19

	Answered by mr W last updated on 10/Aug/19

	$i f a l i n e y = m x + c t a n g e n t s t h e e l l i p s e$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $w e h a v e$ $\frac{x^{2}}{a^{2}} + \frac{{(m x + c)}^{2}}{b^{2}} = 1$ $b^{2} x^{2} + a^{2} (m^{2} x^{2} + 2 m c x + c^{2}) = a^{2} b^{2}$ $(m^{2} a^{2} + b^{2}) x^{2} + 2 {m c a}^{2} x + a^{2} (c^{2} - b^{2}) = 0$ $Δ = 4 m^{2} c^{2} a^{4} - 4 (m^{2} a^{2} + b^{2}) a^{2} (c^{2} - b^{2}) = 0$ $m^{2} c^{2} a^{2} - (m^{2} a^{2} + b^{2}) (c^{2} - b^{2}) = 0$ $\Rightarrow m^{2} a^{2} + b^{2} = c^{2}$ ${l e t}^{'} s s a y a l l t h e t h r e e e l l i p s e s h a v e a$ $c o m m o n t a n g e n t l i n e y = m x + c, t h e n$ $m^{2} a_{1}^{2} + b_{1}^{2} = c^{2} . . . (i)$ $m^{2} a_{2}^{2} + b_{2}^{2} = c^{2} . . . (i i)$ $m^{2} a_{3}^{2} + b_{3}^{2} = c^{2} . . . (i i i)$ $(i) - (i i) :$ $m^{2} (a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = 0 . . . (i v)$ $(i) - (i i i) :$ $m^{2} (a_{1}^{2} - a_{3}^{2}) + (b_{1}^{2} - b_{3}^{2}) = 0 . . . (v)$ $f r o m (i v) a n d (v) :$ $(a_{1}^{2} - a_{3}^{2}) (b_{1}^{2} - b_{2}^{2}) - (a_{1}^{2} - a_{2}^{2}) (b_{1}^{2} - b_{3}^{2}) = 0$ $a_{1}^{2} b_{1}^{2} - a_{3}^{2} b_{1}^{2} - a_{1}^{2} b_{2}^{2} + a_{3}^{2} b_{2}^{2} - a_{1}^{2} b_{1}^{2} + a_{2}^{2} b_{1}^{2} + a_{1}^{2} b_{3}^{2} - a_{2}^{2} b_{3}^{2} = 0$ $\Rightarrow (a_{2}^{2} b_{3}^{2} - a_{3}^{2} b_{2}^{2}) - (a_{1}^{2} b_{3}^{2} - a_{3}^{2} b_{1}^{2}) + (a_{1}^{2} b_{2}^{2} - a_{2}^{2} b_{1}^{2}) = 0$ $\Rightarrow \| \begin{matrix} a_{1}^{2} & b_{1}^{2} & 1 \\ a_{2}^{2} & b_{2}^{2} & 1 \\ a_{3}^{2} & b_{3}^{2} & 1 \end{matrix} \| = 0$
	Commented by peter frank last updated on 10/Aug/19

	$t h a n k y o u v e r y m u c h$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum