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  • If the angle between the asymptotes of hyperbola    <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx

If the angle between the asymptotes of hyperbola    \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}

is  \mathrm{120\degree}  and the product of the perpendiculars drawn from the

foci upon its any tangent is 9, then the locus of the point of

intersection of perpendicular tangents of the hyperbola can be

Option: 1

\mathrm{x^2+y^2=6}


Option: 2

\mathrm{x^2+y^2=9}


Option: 3

\mathrm{x^2+y^2=3}


Option: 4

\mathrm{x^2+y^2=18}


Answers (1)

best_answer

The product of perpendiculars drawn from the foci

upon any of its tangents is 9. Therefore,

         \mathrm{b^2=9}

Also,\mathrm{\frac{b}{a}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}}

\mathrm{\therefore \quad a^2=3 b^2=27}

Therefore, the required locus is the director circle of the

hyperbola which is given by

        \mathrm{x^2+y^2=27-9}

or    \mathrm{x^2+y^2=18}

If  \mathrm{b / a=\tan 60^{\circ}}

\mathrm{a^2=\frac{b^2}{3}=\frac{9}{3}=3}

Hence, the required locus is  \mathrm{x^2+y^2=3-9=-6}  which is

not possible.

Posted by

Pankaj

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