A) 8
B) \[8\sqrt{3}\]
C) \[8\sqrt{2}\]
D) None of these
Correct Answer: C
Solution :
Solving above equation with parabola \[{{y}^{2}}=4x,\]we get the points \[P(3+2\sqrt{2},\,2+2\sqrt{2}),\,\,Q(3-2\sqrt{2},2-2\sqrt{2})\]. \\[P{{Q}^{2}}=32+32=64\]Þ \[PQ=8\] Also, if p be perpendicular from (?1,2) on PQ, then area of triangle is \[\frac{1}{2}PQ.\,\,p=\frac{1}{2}.8.\left( \frac{4}{\sqrt{2}} \right)=8\sqrt{2}\].
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