A) \[(6a,\ -9a)\]
B) \[(-9a,\ 6a)\]
C) \[(-6a,\ 9a)\]
D) \[(9a,\ -6a)\]
Correct Answer: D
Solution :
If normal drawn to point \[(at_{1}^{2},\,2a{{t}_{1}})\] of a parabola \[{{y}^{2}}=4ax\] meets at point \[(at_{2}^{2},\,2a{{t}_{2}})\] of same parabola then, \[{{t}_{2}}=-{{t}_{1}}-2/{{t}_{1}}\] In question \[x=y\] (given) because abscissa and ordinate are equal. \ \[{{y}^{2}}=4ax\]Þ \[{{y}^{2}}=4ay\] [we use relation \[x=y\]] Þ \[{{y}^{2}}=4ay=0\] Þ \[y(y-4a)=0\] Þ \[y=0\] or \[y=4a\] therefore point \[(x=0,\,y=0)\] and \[(x=4a,\,y=4a)\] \[2a{{t}_{1}}=4a\] Þ \[{{t}_{1}}=\frac{4a}{2a}=2\]; \[{{t}_{2}}=-2-\frac{2}{2}=-2-1=-3\] \ \[(at_{2}^{2},\,2a{{t}_{2}})=[a\times {{(-3)}^{2}},\,2a(-3))=(9a,\,-6a)\].You need to login to perform this action.
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