question_answer

Tangents at any point on the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] cut the axes at A and B respectively. If the rectangle OAPB (where O is the origin)is completed, then locus of point P is given by :

A) \[\frac{{{a}^{2}}}{{{x}^{2}}}-\frac{{{b}^{2}}}{{{y}^{2}}}=1\]

B) \[\frac{{{a}^{2}}}{{{x}^{2}}}+\frac{{{b}^{2}}}{{{y}^{2}}}=1\]

C) \[\frac{{{a}^{2}}}{{{y}^{2}}}-\frac{{{b}^{2}}}{{{x}^{2}}}=1\]

D) None of these  

Correct Answer: A

Solution :

[a] Eq. of the tangent at the point \['\theta '\] is

\[\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\]

\[\Rightarrow \] A is \[(a\,cos\theta ,\,\,0)\] and B is \[(0,\,\,-b\,cot\,\,\theta )\]

Let P be \[(h,k)\Rightarrow h=acos\theta ,k=-bcot\theta \]

\[\Rightarrow \frac{k}{h}=-\frac{b}{a\sin \theta }\Rightarrow \sin \theta =\frac{bh}{ak}\] and \[\cos \theta =\frac{h}{a}.\]

Square and add,

\[\Rightarrow \frac{{{b}^{2}}{{h}^{2}}}{{{a}^{2}}{{k}^{2}}}+\frac{{{h}^{2}}}{{{a}^{2}}}=1\Rightarrow \frac{{{b}^{2}}}{{{k}^{2}}}+1=\frac{{{a}^{2}}}{{{h}^{2}}}\]

\[\Rightarrow \frac{{{a}^{2}}}{{{h}^{2}}}-\frac{{{b}^{2}}}{{{k}^{2}}}=1\]

Hence, locus of P is \[\frac{{{a}^{2}}}{{{x}^{2}}}-\frac{{{b}^{2}}}{{{y}^{2}}}=1\]