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Question Number 28278 by ajfour last updated on 23/Jan/18

Commented by ajfour last updated on 23/Jan/18

$I f y = m x + c b e t a n g e n t t o t h e$ $e l l i p s e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, f i n d t h e$ $c o n d i t i o n f o r t a n g e n c y o f t h e$ $g i v e n l i n e t o t h e e l l i p s e .$
	Answered by ajfour last updated on 23/Jan/18

	$x = a \cos θ, y = b \sin θ$ $A s y = m x + c$ $c = b \sin θ - a m \cos θ$ $c = \sqrt{a^{2} m^{2} + b^{2}} \sin (θ - \tan^{- 1} \frac{a m}{b})$ $l i n e c u t s t h e e l l i p s e o n l y i f$ $θ i s r e a l .$ $f o r b e i n g t a n g e n t t h e l i n e h a s$ $g r e a t e s t v a l u e o f c f o r a g i v e n m .$ $h e n c e f o r t h e l i n e y = m x + c t o b e$ $t a n g e n t$ $c = \pm \sqrt{a^{2} m^{2} + b^{2}} .$
	Commented by mrW2 last updated on 23/Jan/18

	$T h a n k y o u s i r!$
	Answered by mrW2 last updated on 23/Jan/18
	$Let′s see the general case and find the condition when the line px+qy+r=0 tangents the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1. The intersection points from the line and the ellipse are the solutions of { ((px+qy+r=0 ...(i))),(((x^2 /a^2 )+(y^2 /b^2 )=1 ...(ii))) :} from (i): qy=−px−r from (ii): b^2 q^2 x^2 +a^2 q^2 y^2 =a^2 b^2 q^2 ⇒b^2 q^2 x^2 +a^2 (−px−r)^2 =a^2 b^2 q^2 ⇒b^2 q^2 x^2 +a^2 (p^2 x^2 +2prx+r^2 )=a^2 b^2 q^2 ⇒(a^2 p^2 +b^2 q^2 )x^2 +2a^2 prx+a^2 (r^2 −b^2 q^2 )=0 When the line only tangents the ellipse, the eqn. above should have one and only one solution. That means Δ=4a^4 p^2 r^2 −4(a^2 p^2 +b^2 q^2 )a^2 (r^2 −b^2 q^2 )=0 ⇒a^2 p^2 r^2 =(a^2 p^2 +b^2 q^2 )(r^2 −b^2 q^2 ) ⇒a^2 p^2 r^2 =a^2 p^2 r^2 +b^2 q^2 r^2 −a^2 b^2 p^2 q^2 −b^4 q^4 ⇒r^2 =a^2 p^2 +b^2 q^2 or ⇒(ap)^2 +(bq)^2 =r^2$
	${L e t}^{'} s s e e t h e g e n e r a l c a s e a n d f i n d$ $t h e c o n d i t i o n w h e n t h e l i n e$ $p x + q y + r = 0$ $t a n g e n t s t h e e l l i p s e$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 .$ $T h e i n t e r s e c t i o n p o i n t s f r o m t h e l i n e$ $a n d t h e e l l i p s e a r e t h e s o l u t i o n s o f$ ${\begin{cases} p x + q y + r = 0 . . . (i) \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (i i) \end{cases}$ $f r o m (i) :$ $q y = - p x - r$ $f r o m (i i) :$ $b^{2} q^{2} x^{2} + a^{2} q^{2} y^{2} = a^{2} b^{2} q^{2}$ $\Rightarrow b^{2} q^{2} x^{2} + a^{2} {(- p x - r)}^{2} = a^{2} b^{2} q^{2}$ $\Rightarrow b^{2} q^{2} x^{2} + a^{2} (p^{2} x^{2} + 2 p r x + r^{2}) = a^{2} b^{2} q^{2}$ $\Rightarrow (a^{2} p^{2} + b^{2} q^{2}) x^{2} + 2 a^{2} p r x + a^{2} (r^{2} - b^{2} q^{2}) = 0$ $W h e n t h e l i n e o n l y t a n g e n t s t h e e l l i p s e,$ $t h e e q n . a b o v e s h o u l d h a v e o n e a n d$ $o n l y o n e s o l u t i o n . T h a t m e a n s$ $Δ = 4 a^{4} p^{2} r^{2} - 4 (a^{2} p^{2} + b^{2} q^{2}) a^{2} (r^{2} - b^{2} q^{2}) = 0$ $\Rightarrow a^{2} p^{2} r^{2} = (a^{2} p^{2} + b^{2} q^{2}) (r^{2} - b^{2} q^{2})$ $\Rightarrow a^{2} p^{2} r^{2} = a^{2} p^{2} r^{2} + b^{2} q^{2} r^{2} - a^{2} b^{2} p^{2} q^{2} - b^{4} q^{4}$ $\Rightarrow r^{2} = a^{2} p^{2} + b^{2} q^{2}$ $o r$ $\Rightarrow {(a p)}^{2} + {(b q)}^{2} = r^{2}$
	Commented by ajfour last updated on 23/Jan/18

	$t h a n k s S i r, n i c e g e n e r a l i s a t i o n .$
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