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  • The locus of the foot of the perpendicular from the centre of the hyperbola xy=1 on a variable tangent isOption: 1 <img alt="\mathrm{\left(

The locus of the foot of the perpendicular from the centre of the

hyperbola xy=1 on a variable tangent is

Option: 1

\mathrm{\left(x^2-y^2\right)^2=4 x y}


Option: 2

\mathrm{\left(x^2+y^2\right)^2=2 x y}


Option: 3

\mathrm{\left(x^2+y^2\right)^2=4 x y}


Option: 4

None of these


Answers (1)

best_answer

Let the foot perpendicular from O(0,0) to the tangent

to the hyperbola be P(h,k).

Slope of  \mathrm{\mathrm{OP}=\frac{k}{h}}

Then the equation of the tangent to the hyperbola is

\mathrm{\begin{aligned} & y-k=-\frac{h}{k}(x-h) \\ & h x+k y=h^2+k^2 \end{aligned}}

Solving it with xy=1, we have

          \mathrm{h x+\frac{k}{x}=h^2+k^2}

or   \mathrm{h x^2-\left(h^2+k^2\right) x+k=0}

This equation must have real and equal roots.Hence,

               \mathrm{ D=0}

Or   \mathrm{\left(h^2+k^2\right)^2-4 h k=0}

Or    \mathrm{\left(x^2+y^2\right)^2=4 x y}

 

Posted by

Irshad Anwar

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