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I read the following property about the condition for a point such that two tangents can be drawn from it to a hyperbola, somewhere:

We determine this by simply making asymptotes from the centre of the hyperbola. This divides the plane into four regions. Now,

1) If your external point lies in the region in which the first branch lies, both the tangents will be drawn on the second branch.

2)If your external point lies in the region in which the second branch lies, both the tangents will be drawn on the first branch.

3)If your external point lies in the two regions in which no branch lies, one tangent will be drawn on the first branch and one tangent will be drawn on the second branch.

I was thinking about proving it but was not able to do anything. Need help.

EditPiAf
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Raghav Chaturvedi
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  • Certainly, for a point on the hyperbola itself, the (double) tangent meets the branch in the same region as the point. For a point slightly exterior to the hyperbola, one still expects that the tangents to meet the branch in the same region as the point. (Consider, specifically, a point moving from one vertex to the other, the tangents will meet the "same-region" branch up until the point hits the center, where the "tangents" are the asymptotes; beyond there, the tangents meet the other branch, but then the point is also in that other branch's region.) Your (1) and (2) are incorrect. – Blue May 13 '17 at 16:18

1 Answers1

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By using an affine transformation you can take any hyperbola to a standard one, so we can just prove it for one hyperbola. I suggest using $xy=1$. That makes the asymptotes the coordinate axes and the regions the four quadrants of the plane. Now just take an arbitrary point $(a,b)$, find the tangents, and show the properties you have claimed depending on the signs of $a,b$. Everything is symmetric on rotation by $\pi$, so you only have to consider $b \ge 0$. I believe if $a=0$ or $b=0$ there are no tangents at all.

Ross Millikan
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