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The angle between the asymptotes of hyperbola  \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is 600 and the product of perpendicular  drawn from the foci upon its any tangent is 9, then locus of point of intersection of perpendicular tangents of the hyperbola can be

Option: 1

x^2-y^2=9


Option: 2

x^2+y^2=9


Option: 3

x^2-y^2=18


Option: 4

x^2+y^2=18


Answers (1)

 

 

Asymptotes of Hyperbola -

 

TIP:

The angle between the asymptotes of the hyperbola \frac{y^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { is } 2 \tan ^{-1}\left(\frac{b}{a}\right)

If the angle between the asymptotes of the hyperbola { \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { is } 2 \theta \text { then } e=\sec \theta}

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Angle between asymptotes

 \\2\tan^{-1}\frac{b}{a}=60^0\\\tan^{-1}\frac{b}{a}=30^0\\\frac{b}{a}=\frac{1}{\sqrt 3}\\a^2=3b^2 

\\\because b^2=9\\\Rightarrow a^2=27

\therefore\text{required locus is }x^2+y^2=27-9=18

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Kshitij

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