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Question Number 83262 by peter frank last updated on 29/Feb/20

	Answered by mr W last updated on 29/Feb/20

	$c u r v e 1 : x^{2} + 4 y^{2} = 8 . . . (i)$ $c u r v e 2 : x^{2} - 2 y^{2} = 4 . . . (i i)$ $i n t e r s e c t i o n :$ $(i) - (i i) :$ $6 y^{2} = 4 \Rightarrow y = \pm \frac{2}{\sqrt{6}}$ $(i) + (i i) \times 2 :$ $3 x^{2} = 16 \Rightarrow x = \pm \frac{4}{\sqrt{3}}$ $t a n g e n t o f c u r v e 1 a t i n t e r s e c t i o n :$ $2 x + 8 {y y}^{'} = 0$ $\Rightarrow \tan θ_{1} = y^{'} = - \frac{1}{4} (\frac{x}{y}) = - \frac{1}{4} (\pm 2 \sqrt{2}) = \mp \frac{\sqrt{2}}{2}$ $t a n g e n t o f c u r v e 2 a t i n t e r s e c t i o n :$ $2 x - 4 {y y}^{'} = 0$ $\Rightarrow \tan θ_{2} = y^{'} = \frac{1}{2} (\frac{x}{y}) = \frac{1}{2} (\pm 2 \sqrt{2}) = \pm \sqrt{2}$ $\tan (θ_{2} - θ_{1}) = \frac{\pm \sqrt{2} - (\mp \frac{\sqrt{2}}{2})}{1 + (\pm \sqrt{2}) (\mp \frac{\sqrt{2}}{2})} = \pm \infty$ $∣ θ_{2} - θ_{1} ∣= \frac{π}{2}$
	Commented by peter frank last updated on 29/Feb/20

	$t h a n k y o u$
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