SATHEE: Parabola 1 Question 1

Parabola 1 Question 1

1. If the area of the triangle whose one vertex is at the vertex of the parabola, $y^{2} + 4 (x - a^{2}) = 0$ 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

(d) $(10)^{2 / 3}$

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Answer:

Correct Answer: 1. (b)

Solution:

Vertex of parabola $y^{2} = - 4 (x - a^{2})$ is $(a^{2}, 0)$ .

For point of intersection with $Y$ -axis, put $x = 0$ in the given equation of parabola.

This gives, $y^{2} = 4 a^{2} \Rightarrow y = \pm 2 a$

Thus, the point of intersection are $(0, 2 a)$ and $(0, - 2 a)$ .

From the given condition, we have

Area of $△ A B C = 250$

$\begin{aligned} ∴ \frac{1}{2} (B C) (O A) = 250 [∵ Area = \frac{1}{2} \times base \times height] \\ \begin{aligned} \Rightarrow & \frac{1}{2} (4 a) a^{2} & = 250 \\ ∴ & a & = 5 \end{aligned} \end{aligned}$

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Parabola 1 Question 1

1. If the area of the triangle whose one vertex is at the vertex of the parabola, y2+4(x−a2)=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

Answer:

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1. If the area of the triangle whose one vertex is at the vertex of the parabola, $y^{2} + 4 (x - a^{2}) = 0$ 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