A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
We have, \[\frac{2}{{{t}_{1}}}.\,\frac{2}{{{t}_{2}}}\,=-1\,\Rightarrow \,{{t}_{1}}{{t}_{2}}\,=-4;\] Also, \[{{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\] \[{{t}_{1}}{{t}_{2}}\,=-t_{1}^{2}-2\Rightarrow \,-4+2=-t_{1}^{2}\Rightarrow \,t_{1}^{2}=2\] Also \[t_{2}^{2}\,=t_{1}^{2}\,+\frac{4}{t_{1}^{2}}+4\,\] squaring (1) \[\Rightarrow \,\,t_{2}^{2}=2+2+4=8\] \[SQ=a(1+t_{2}^{2})\,=a(1+8)=9a\] \[SP=a(1+t_{1}^{2}\,)=a(1+2)=3a\] \[\therefore \,\,\frac{SQ}{SP}=3\Rightarrow \,SQ=3SP\Rightarrow \,\lambda =3\]You need to login to perform this action.
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