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If \mathrm{PN} is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, the mid-point of \mathrm{PN} is 
 

Option: 1

Circle


Option: 2

Parabola


Option: 3

Ellipse


Option: 4

Hyperbola


Answers (1)

best_answer

Let \mathrm{xy=c^{2}} be the rectangular hyperbola, and let \mathrm{P(x_{1},y_{1})} be a point on it. Let \mathrm{Q(h,k)} be the mid-point of \mathrm{PN}. Then the coordinates of \mathrm{Q} are \mathrm{\left(x_1, \frac{y_1}{2}\right)}
\mathrm{\therefore x_1=h \text { and } \frac{y_1}{2}=k \Rightarrow x_1=h \text { and } y_1=2 k}
But \mathrm{(x_{1},y_{1})} lies on \mathrm{xy=c^{2}}
\mathrm{\therefore \quad h \cdot(2 k)=c^2 \Rightarrow h k \Rightarrow c^2 / 2}
Hence, the locus of \mathrm{(h,k)} is \mathrm{x y=c^2 / 2}, which is a hyperbola.

Posted by

Kuldeep Maurya

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