Conic Sections Question 192
Question: 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]
Options:
A) $ (6a,\ -9a) $
B) $ (-9a,\ 6a) $
C) $ (-6a,\ 9a) $
D) $ (9a,\ -6a) $
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Answer:
Correct Answer: D
Solution:
If normal drawn to point $ (at_1^{2},2at_1) $ of a parabola $ y^{2}=4ax $ meets at point $ (at_2^{2},2at_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 $
therefore $ y^{2}=4ay $
[we use relation $ x=y $ ]
therefore $ y^{2}=4ay=0 $
therefore $ y(y-4a)=0 $
therefore $ y=0 $ or $ y=4a $
therefore point $ (x=0,y=0) $ and $ (x=4a,y=4a) $
$ 2at_1=4a $
therefore $ t_1=\frac{4a}{2a}=2 $ ; $ t_2=-2-\frac{2}{2}=-2-1=-3 $
$ (at_2^{2},2at_2)=[a\times {{(-3)}^{2}},2a(-3))=(9a,-6a) $ .
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