Question:
<p>If the area of the triangle whose one vertex is at the vertex of the parabola, y<sup>2</sup> + 4(x - a<sup>2</sup>) = 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</p>
<p style="display:inline">5(2<sup>1/3</sup>)</p>
<p style="display:inline">(10)<sup>2/3</sup></p>
<p style="display:inline">5<span class="math-tex">\(\sqrt5\)</span></p>
<p style="display:inline">5</p>
Step-by-Step Solution
Key Concept: Identify the parabola's vertex and y-intercepts to determine the triangle's base and height for the area equation.
<p>Vertex of parabola y<sup>2 </sup>= -4(x - a<sup>2</sup>) is (a<sup>2</sup>, O).<br />
For point of intersection with Y-axis, put x = 0 in the<br />
given equation of parabola.<br />
This gives, y<sup>2</sup> = 4a<sup>2</sup> <span class="math-tex">$\Rightarrow$</span> y = ± 2a<br />
Thus, the point of intersection are (0, 2a) and (0, -2a).<br />
<img alt="" data-imgur-src="PATBeon.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/PATBeon.png" style="width: 200px; height: 164px;" /><br />
From the given condition, we have<br />
Area of <span class="math-tex">$\Delta$</span>ABC = 250<br />
<span class="math-tex">$\therefore \frac{1}{2}(B C)(O A)=250 \quad\left[\because \text { Area }=\frac{1}{2} \times \text { base } \times \text { height }\right]$</span><br />
<span class="math-tex">$\Rightarrow \quad \frac{1}{2}(4 a) a^{2}=250 \quad \Rightarrow a^{3}=125=5^{3}$</span><br />
<span class="math-tex">$\therefore$</span> a = 5</p>
Correct Answer: D