Prove that the perpendicilar tangent to the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 meets on the circle x^2 +y^2 =a^2 +b^2 .


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Question Number 51613 by peter frank last updated on 29/Dec/18

$P r o v e t h a t t h e p e r p e n d i c i l a r$ $t a n g e n t t o t h e e l l i p s e$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 m e e t s o n t h e$ $c i r c l e x^{2} + y^{2} = a^{2} + b^{2} .$
	Answered by peter frank last updated on 29/Dec/18

	$y = m x + \sqrt{a^{2} m^{2} + b^{2}}$ $y - m x = \sqrt{a^{2} m^{2} + b^{2}}$ ${(y - m x)}^{2} = {(\sqrt{a^{2} m^{2} + b^{2}})}^{2}$ $m^{2} (x^{2} - a^{2}) - 2 m x y + y^{2} - b^{2} = 0$ $m_{1} + m_{2} = \frac{2 m x}{x^{2} - a^{2}}$ $m_{1} m_{2} = \frac{y^{2} - b^{2}}{x^{2} - a^{2}}$ $f o r p e r p e n d i c u l a r$ $m_{1} m_{2} = - 1$ $- 1 = \frac{y^{2} - b^{2}}{x^{2} - a^{2}}$ $y^{2} + x^{2} = a^{2} + b^{2}$
	Answered by mr W last updated on 29/Dec/18

	$t w o p e r p e n d i c u l a r l i n e s f r o m p o i n t$ $(h, k) :$ $\frac{y - k}{x - h} = m \Rightarrow m x - y - m h + k = 0$ $\frac{y - k}{x - h} = - \frac{1}{m} \Rightarrow x + m y - h + m k = 0$ $s i n c e t h e y t a n g e n t t h e e l l i p s e,$ $a^{2} m^{2} + b^{2} {(- 1)}^{2} - {(- m h + k)}^{2} = 0$ $\Rightarrow a^{2} m^{2} + b^{2} = m^{2} h^{2} + k^{2} - 2 m h k . . . (i)$ $a^{2} {(1)}^{2} + b^{2} m^{2} - {(- h + m k)}^{2} = 0$ $\Rightarrow a^{2} + b^{2} m^{2} = h^{2} + m^{2} k^{2} - 2 m h k . . . (i i)$ $(i) + (i i) :$ $\Rightarrow a^{2} (m^{2} + 1) + b^{2} (1 + m^{2}) = h^{2} (m^{2} + 1) + k^{2} (1 + m^{2})$ $\Rightarrow a^{2} + b^{2} = h^{2} + k^{2}$ $o r$ $\Rightarrow x^{2} + y^{2} = a^{2} + b^{2}$
	Commented by peter frank last updated on 29/Dec/18

	$t h a n k y o u$
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