Prove that line lx+my+n=0 is tangent to the ellipse (x^2 /a^2 )+(y^2 /b^(2 ) )=1 if a^2 l^2 +b^2 m^2 =n^2


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Question Number 77127 by peter frank last updated on 03/Jan/20

$P r o v e t h a t l i n e l x + m y + n = 0$ $i s t a n g e n t t o t h e e l l i p s e$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 i f a^{2} l^{2} + b^{2} m^{2} = n^{2}$
	Answered by jagoll last updated on 03/Jan/20

	$s u p p o s e a > b \Leftrightarrow t a n g e n t l i n e$ $e l l i p s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow y = m x \pm \sqrt{a^{2} m^{2} + b^{2}}$ $e q u a l t o l x + m y + n = 0 \Rightarrow$ $y = - \frac{l}{m} x - \frac{n}{m}$ $m = - \frac{l}{m} \Rightarrow m^{2} = - l$ $- \frac{n}{m} = - \sqrt{a^{2} m^{2} + b^{2}}$ $\Rightarrow \frac{n^{2}}{m^{2}} = a^{2} m^{2} + b^{2}$ $n^{2} = (- l) (- {l a}^{2} + b^{2})$ $n^{2} = l^{2} a^{2} - {l b}^{2} \Rightarrow n^{2} = l^{2} a^{2} + m^{2} b^{2}$
	Commented by peter frank last updated on 03/Jan/20

	$t h a n k y o u$
	Answered by mr W last updated on 03/Jan/20

	$b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$ $m^{2} b^{2} x^{2} + a^{2} m^{2} y^{2} = m^{2} a^{2} b^{2}$ $m^{2} b^{2} x^{2} + a^{2} {(l x + n)}^{2} = m^{2} a^{2} b^{2}$ $m^{2} b^{2} x^{2} + a^{2} (l^{2} x^{2} + n^{2} + 2 l n x) = m^{2} a^{2} b^{2}$ $(a^{2} l^{2} + b^{2} m^{2}) x^{2} + 2 a^{2} l n x + a^{2} (n^{2} - b^{2} m^{2}) = 0$ $d u e t o t a n g e n c y :$ $Δ = {(2 a^{2} l n)}^{2} - 4 (a^{2} l^{2} + b^{2} m^{2}) a^{2} (n^{2} - b^{2} m^{2}) = 0$ $a^{2} l^{2} n^{2} - (a^{2} l^{2} + b^{2} m^{2}) (n^{2} - b^{2} m^{2}) = 0$ $(- n^{2} + a^{2} l^{2} + b^{2} m^{2}) b^{2} m^{2} = 0$ $\Rightarrow a^{2} l^{2} + b^{2} m^{2} = n^{2}$
	Commented by peter frank last updated on 03/Jan/20

	$t h a n k y o u$
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