A) \[{{p}^{2}}+pq+2=0\]
B) \[{{p}^{2}}-pq+2=0\]
C) \[{{q}^{2}}+pq+2=0\]
D) \[{{p}^{2}}+pq+1=0\]
Correct Answer: A
Solution :
We know that the normal drawn at a point\[P(at_{1}^{2},\,\,2a{{t}_{1}})\] to the parabola\[{{y}^{2}}=4ax\]meets again the parabola at \[Q\,(at_{2}^{2},2a{{t}_{2}}),\]then \[{{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\] Here, \[{{t}_{1}}=p\] and \[{{t}_{2}}=q\] \[\therefore \] \[q=-p-\frac{2}{p}\] \[\Rightarrow \] \[pq=-{{p}^{2}}-2\] \[\Rightarrow \] \[{{p}^{2}}+pq+2=0.\]
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