A) \[6\sqrt{3}\]
B) \[3\sqrt{3}\]
C) 2
D) 1
Correct Answer: A
Solution :
Normal at \[P(t_{1}^{2},\,2{{t}_{1}})\] on the parabola \[{{y}^{2}}=4x\] .....(i) Meets it again at the point \[(y-12)=\frac{-36}{54}(x+36)\Rightarrow 2x+3y+36=0\], where \[{{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\] .....(ii) If \[PQ\] subtends a right angle at the vertex (0, 0) then (Slope of OP) (Slope of \[OQ)\] \[=-1\] \[\Rightarrow \,\]\[\frac{2{{t}_{1}}}{t_{1}^{2}}.\,\frac{2{{t}_{2}}}{t_{2}^{2}}=-1\]\[\Rightarrow \]\[{{t}_{2}}=-\frac{4}{{{t}_{1}}}\] .....(iii) From (ii) and (iii), \[-{{t}_{1}}-\frac{2}{{{t}_{1}}}=-\frac{4}{{{t}_{1}}}\]\[\Rightarrow \]\[-{{t}_{1}}=-\frac{2}{{{t}_{1}}}\] \[\Rightarrow \,t_{1}^{2}\]= 2 \[\Rightarrow \,\,{{t}_{1}}=\pm \,\sqrt{2}\]; \[\therefore \,\,{{t}_{2}}=\mp \,2\,\sqrt{2}\] \[\therefore \,\]\[P\] and \[Q\] are \[(2,\,\pm \,2\sqrt{2})\] and \[(8,\,\mp \,4\sqrt{2})\] \[\therefore \,\]\[PQ=\sqrt{{{(8-2)}^{2}}+{{(\mp \,4\sqrt{2}\mp 2\sqrt{2})}^{2}}}=\sqrt{36+72}\] \[=\sqrt{108}=6\sqrt{3.}\]
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