question_answer

The normal meet the parabola \[{{y}^{2}}=4ax\] at that point where the abissiae of the point is equal to the ordinate of the point is [DCE 2005]

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 \[{{t}_{1}}=\frac{4a}{2a}=2\]; \[{{t}_{2}}=-2-\frac{2}{2}=-2-1=-3\]                    \ \[(at_{2}^{2},\,2a{{t}_{2}})=[a\times {{(-3)}^{2}},\,2a(-3))=(9a,\,-6a)\].