JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the area of the triangle whose one vertex is at the vertex of the parabola, \({y^2} + 4\,\left( {x - {a^2}} \right) = 0\) and the other two vertices are the points of intersection of the parabola and \(y -\) axis, is \(250\, sq\). units, then a value of \(‘a’\) is
- A \(5\sqrt 5 \)
- B \(5\left( {{2^{1/3}}} \right)\)
- C \({\left( {10} \right)^{2/3}}\)
- D \(5\)
Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
\({y^2} = - 4\left( {x - {a^2}} \right)\) Vertices of triangle are \(\left( {{a^2},0} \right)\) and \(\left( {0,2a} \right)\) and \(\left( {0, - 2a} \right)\) Area \( = \frac{1}{2}\left( {{a^2}} \right)\left( {4a} \right) = 250\) \( \Rightarrow {a^3} = 125\)
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