A) 1
B) 2
C) 3
D) 0
Correct Answer: D
Solution :
Let the point be\[P\left( ct,\,\,\frac{c}{t} \right)\] The equation of the tangent to the hyperbola \[xy={{c}^{2}}\]at\[P\]is \[x\frac{c}{t}+cty=2{{c}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,x+y{{t}^{2}}-2ct=0\] ?(i) Then, putting\[y=0,\,\,x=0\]successively in Eq. (i), we get\[{{a}_{1}}=2ct\]and\[{{b}_{1}}=\frac{2c}{t}\] (\[\because \]intercepts of tangent on the axes are\[{{a}_{1}}\]and \[{{b}_{1}}\]) Again the equation of the normal to the hyperbola\[xy={{c}^{2}}\]at\[P\]is \[x{{t}^{3}}-yt-c{{t}^{4}}+c=0\] As before,\[{{a}_{2}}=ct-\frac{c}{{{t}^{3}}}\]and\[{{b}_{2}}=-c{{t}^{3}}+\frac{c}{t}\] \[\therefore \,{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=2ct\,\left( ct\,-\frac{c}{{{t}^{3}}} \right)\,+\frac{2c}{t}\,\left( -c{{t}^{3}}+\frac{c}{t} \right)\] \[=0\]
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