question_answer

A tangent to the hyperbola \[{{x}^{2}}-2{{y}^{2}}=4\] meets x-axis at P and y-axis at Q. Line PR and QR are drawn such that OPRQ is a rectangle (where 0 is origin). The locus of R is

A)  \[\frac{4}{{{x}^{2}}}+\frac{2}{{{y}^{2}}}=1\]                       

B)  \[\frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=1\]

C)  \[\frac{2}{{{x}^{2}}}-\frac{4}{{{y}^{2}}}=1\]                        

D)  \[\frac{2}{{{x}^{2}}}+\frac{4}{{{y}^{2}}}=1\]

Correct Answer: B

Solution :

\[\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1\]                 \[\therefore P(a\cos \theta ,\,\,0)\,\,Q(0,\,\,-b\cot \theta )\] \[\therefore h=a\cos \theta \]  \[\Rightarrow \sec \theta =\frac{a}{h}\] \[k=-b\cot \theta \]        \[\Rightarrow \tan \theta =\frac{-b}{k}\] So, \[\frac{{{a}^{2}}}{{{x}^{2}}}-\frac{{{b}^{2}}}{{{y}^{2}}}=1\Rightarrow \frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=1\]