A) \[5\sqrt{5}\]
B) \[{{(10)}^{2/3}}\]
C) \[5{{(2)}^{1/3}}\]
D) 5
Correct Answer: D
Solution :
The vertex of the given parabola is\[({{a}^{2}},0).\] When \[x=0,{{y}^{2}}+4(0-{{a}^{2}})=0\Rightarrow y=\pm 2a\] The point of intersection of the given parabola and the y-axis are \[\left( 0,\pm 2a \right).\] \[\therefore \]Area of the triangle \[=\frac{1}{2}.4a.{{a}^{2}}\] \[\Rightarrow \]\[250=2{{a}^{3}}\Rightarrow a=5\]You need to login to perform this action.
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