If C is the centre of the rectangular hyperbola, then the product of the slopes of CP, CQ, CR and CS is equal to
A) -1
B) 1
C) 0
D) None of these
Correct Answer: B
Solution :
Let \[{{t}_{1}},{{t}_{2}},{{t}_{3}},{{t}_{4}}\] be the parameters of the points P,Q,R and S respectively. Then, the coordinates of P, Q, R and S are \[\left( c{{t}_{1}},\frac{c}{{{t}_{1}}} \right),\left( c{{t}_{2}},\frac{c}{{{t}_{2}}} \right),\left( c{{t}_{3}},\frac{c}{{{t}_{3}}} \right)\]and\[\left( c{{t}_{4}},\frac{c}{{{t}_{4}}} \right)\]respectively. Since, \[PQ\bot RS\] \[\frac{\frac{c}{{{t}_{2}}}-\frac{c}{{{t}_{1}}}}{c{{t}_{2}}-c{{t}_{1}}}\times \frac{\frac{c}{{{t}_{4}}}-\frac{c}{{{t}_{3}}}}{c{{t}_{4}}-c{{t}_{3}}}=-1\] \[\Rightarrow \]\[-\frac{1}{{{t}_{1}}{{t}_{2}}}\times \left( \frac{1}{{{t}_{3}}{{t}_{4}}} \right)=-1\]\[\Rightarrow \]\[{{t}_{1}}{{t}_{2}}{{t}_{3}}{{t}_{4}}=-1\] ?(i) Slope of \[CP=\frac{\frac{c}{{{t}_{1}}}-0}{c{{t}_{1}}-0}=\frac{1}{t_{1}^{2}}\] Similarly, slopes of CQ, CR and CS are \[\frac{1}{t_{2}^{2}},\frac{1}{t_{3}^{2}}\] and\[\frac{1}{t_{4}^{2}}\] respectively. \[\therefore \]Product of die slopes of CP, CQ, CR and CS \[=\frac{1}{t_{1}^{2}}\times \frac{1}{t_{2}^{2}}\times \frac{1}{t_{3}^{2}}\times \frac{1}{t_{4}^{2}}=\frac{1}{t_{1}^{2}t_{2}^{2}t_{3}^{2}t_{4}^{2}}=1\][From Eq. (i)]
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