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For the hyperbola, $\frac{x^2}{16}-\frac{y^2}{25}=1$, director circle is $x^2+y^2=-9$, which is not possible. Does that mean director circle is an optional feature of the hyperbola?

Edit: Director circle is the locus of point of intersection of perpendicular tangents. Its equation for the hyperbola is $x^2+y^2=a^2-b^2$.

aarbee
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  • I am not able to type exponent somehow. Can someone edit please? – aarbee Dec 29 '20 at 16:59
  • I put exponents in. Please check it's what you mean. Also what is "director circle"? – coffeemath Dec 29 '20 at 17:02
  • To put an exponent in use ^ -- x^2 inside the dollar signs – postmortes Dec 29 '20 at 17:06
  • @coffeemath Thanks. I have added the definition of director circle. It would be nice if you edit that too. Thanks – aarbee Dec 29 '20 at 17:06
  • @postmortes, yes, but I am not able to use that key somehow. – aarbee Dec 29 '20 at 17:07
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    From Wikipedia: "The director circle of a hyperbola has radius $\sqrt{a^{2} - b^{2}}$, and so, may not exist in the Euclidean plane, but could be a circle with imaginary radius in the complex plane." (https://en.wikipedia.org/wiki/Director_circle) – Joshua Wang Dec 29 '20 at 17:19
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    @aarbee: (For a hyperbola w/ horizontal transverse axis) Think about how a hyperbola's asymptotes restrict the range of tangent slopes, and thus also the range of angles between pairs of tangents. Then consider how $a$ & $b$ affect things: when $a>b$, the asymptotes are "shallow"; at $a=b$, they're perpendicular; for $a<b$, they're "steep". Under only one of those conditions does it make sense that two tangents can be perpendicular. (Well, another works if we treat the asymptotes themselves as tangents to pts "at infinity". The resulting circle eqn reflects the special nature of this case.) – Blue Dec 29 '20 at 17:51

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