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  • If the area of the triangle whose one vertex is at the vetex of the  parabola,  <img alt="\mathrm {y^2+4\left(x-a^2\right)=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5

If the area of the triangle whose one vertex is at the vetex of the  parabola,  \mathrm {y^2+4\left(x-a^2\right)=0}  and the other two vertices are the points of intersection of the parabola and  y-axis, is 250 sq. units, then a value of \mathrm {a} is
 

Option: 1

5 \sqrt{5}


Option: 2

(10)^{2 / 3}


Option: 3

5\left(2^{1 / 3}\right)


Option: 4

5


Answers (1)

best_answer

The vertex of the given parabola is  \mathrm {\left(a^2, 0\right).}
When  \mathrm {x=0, y^2+4\left(0-a^2\right)=0 \Rightarrow y= \pm 2 a} 
The point of intersection of the given parabola and the y-axis are \mathrm {(0, \pm 2 a).} 

 \therefore Area of the triangle =\mathrm {\frac{1}{2} \cdot 4 a \cdot a^2} 

\mathrm {{\Rightarrow \quad 250=2 a^3 \Rightarrow a=5}}

 

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Nehul

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