Parabola Ques 13
- 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
(2019 Main, 11 Jan, II)
(a) $5 \sqrt{5}$
(b) 5
(c) $5\left(2^{1 / 3}\right)$
(d) $(10)^{2 / 3}$
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Answer:
Correct Answer: 13.(b)
Solution:
- Vertex of parabola $y^{2}=-4\left(x-a^{2}\right)$ is $\left(a^{2}, 0\right)$.
For point of intersection with $Y$-axis, put $x=0$ in the given equation of parabola.
This gives, $y^{2}=4 a^{2} \Rightarrow \quad y= \pm 2 a$
Thus, the point of intersection are $(0,2 a)$ and $(0,-2 a)$.
From the given condition, we have
Area of $\triangle A B C=250$
$$ \begin{aligned} & \therefore \quad \frac{1}{2}(B C)(O A)=250 \quad\left[\because \text { Area }=\frac{1}{2} \times \text { base } \times \text { height }\right] \\ & \begin{aligned} & \Rightarrow & \frac{1}{2}(4 a) a^{2} & =250 \\ & \therefore & a & =5 \end{aligned} \end{aligned} $$
BETA
