A) \[9{{y}^{2}}=4x\]
B) \[9{{y}^{2}}=-4x\]
C) \[3{{y}^{2}}=2x\]
D) \[3{{y}^{2}}=-2x\]
Correct Answer: B
Solution :
End points of double ordinate can be taken as \[\left( -{{t}^{2}},2t \right)\And \left( -{{t}^{2}},-2t \right)\] according to given condition. \[x=\frac{-2{{t}^{2}}-{{t}^{2}}}{3}\And y=\frac{-4t+2t}{3}\] \[\Rightarrow \]\[3x=-3{{t}^{2}}\And 3y=-2t\] i.e., \[x=-{{t}^{2}}\And t=-{}^{3}/{}_{2}y\]eliminatry t \[x=-{{\left( -\frac{3y}{2} \right)}^{2}}\]i.e., \[x=-\frac{9}{4}{{y}^{2}}\]i.e., \[9{{y}^{2}}=-4x\]You need to login to perform this action.
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