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Find all the affine tangents that are simultaneously tangent to the set $E$ and $H$: $$H=\{(x,y)\in \mathbb R^2:xy=-5\}, E=\{(x,y)\in \mathbb R^2:\frac{x^2}{9}+\frac{y^2}{4}=1\}$$

I know that when the line is tangent to the set A at point a: $T_aA=\{v:<grad(h(a)),v>=0\}$ but I don't know how to combine this condition for both sets $ E, H $ to find all tangents

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Any tangent to the given ellipse is $$y=mx \pm \sqrt{9m^2+4}$$, insert this in the Eq. of hyperbola:$xy=-5$, to get $$mx^2+\pm\sqrt{9m^2+4}x+5=0$$ For tangency we impose $B^2=4AC$, Then $$9m^2+4=20m \implies m=2, 2/9$$ The tangents are $$y=2x\pm 2\sqrt{10},~~ y=2x/9\pm2\sqrt{10}/3.$$

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Z Ahmed
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  • @kp123 For any real value of $m$, it is tangent to the given ellipse. – Z Ahmed Feb 03 '21 at 17:39
  • It is well know: let $y=mx+c$ be the tangent to x^2/a^2+y^2/b^2=1, put the line in ellipse to get a quadratic of $x$. There you demand single root by imposing $B^2=4AC$ you get $c=\pm \sqrt{a^2m^2+b^2}$. Hence $y=mx \pm \sqrt{a^2m^2+b^2}$ always tangent to this ellipse for any real value of $m$. – Z Ahmed Feb 03 '21 at 17:50