Problem 111 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

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) \((10)^{23}\) (d) 5

Step-by-Step Solution

Answer

The value of \(a\) is 5.

Step 1: Identify the Parabola Equation

The given parabolic equation is: \( y^2 + 4(x - a^2) = 0 \). Rewriting it gives: \( y^2 = -4(x - a^2) \). This equation represents a parabola that opens leftward.

Step 2: Determine the Vertex of the Parabola

The vertex of the parabola is given as \((a^2, 0)\). This is derived directly from the standard form of the parabola \( y^2 = -4(x - h) \), where \( h = a^2 \).

Step 3: Find the Points of Intersection with the Y-Axis

For the points of intersection with the y-axis, set \( x = 0 \). Thus, \( y^2 = -4(0 - a^2) = 4a^2 \). So, the points of intersection are \((0, 2a)\) and \((0, -2a)\).

Step 4: Calculate the Area of the Triangle

The other vertices of the triangle are on the y-axis at \((0, 2a)\) and \((0, -2a)\), with the vertex \((a^2, 0)\). The base of the triangle on the y-axis is \( (2a) - (-2a) = 4a \). The height is the distance from \((a^2,0)\) to the y-axis, which is \(a^2\). The area \(A\) is given by the formula for the area of a triangle: \(A = \frac{1}{2} \times \text{base} \times \text{height}\). Substitute in the known values: \(A = \frac{1}{2} \times 4a \times a^2 = 2a^3 \).

Step 5: Solve for \(a\)

The area of the triangle is given as 250 square units. Set up the equation: \(2a^3 = 250\). Solving gives \(a^3 = 125\). Taking the cube root on both sides, \(a = 5\).

Key Concepts

Parabola Equation

A parabola is a curve described by a specific type of quadratic equation. Understanding the equation of a parabola helps to determine its direction, vertex, and other characteristics. The equation provided in the exercise is \( y^2 + 4(x - a^2) = 0 \). This can be rewritten as \( y^2 = -4(x - a^2) \), revealing its nature.

This form is known as the standard equation of a parabola that opens horizontally. When the equation is expressed as \( y^2 = -4p(x - h) \), it indicates a parabola that opens leftward, since the coefficient of \( x \) is negative. Here, the value of \( h \) is critical because it specifies the horizontal shift of the parabola from the origin.

By recognizing the structure, we can determine key aspects of the parabola's shape and orientation. In this instance, the equation reveals that the parabola opens to the left with its vertex located on the positive axis offset by \( a^2 \) from the origin.

Area of Triangle

Calculating the area of a triangle is an essential geometric task, and it requires knowledge of specific dimensions. For triangles with a vertical or horizontal side, calculations become simpler. The area \( A \) can be calculated when the base and height are known.

In this case, the base of the triangle is the vertical distance between the points \((0, 2a)\) and \((0, -2a)\), calculated as \((2a) - (-2a) = 4a\). The height is the horizontal distance from the vertex of the parabola \((a^2, 0)\) to the y-axis, simply \(a^2\).

Using the standard area formula for a triangle, \( A = \frac{1}{2} \times \text{base} \times \text{height} \), we substitute the base and height. This results in:

• Base = \(4a\)
• Height = \(a^2\)

Thus, the area becomes \( A = \frac{1}{2} \times 4a \times a^2 = 2a^3 \). Plugging in the known area, 250, provides the solution needed to determine \( a \).

Vertices of Parabola

In any parabolic equation, identifying the vertex is crucial as it represents the peak or lowest point of the parabola. The vertex of a horizontally opening parabola, such as the one described by \( y^2 = -4(x - a^2) \), can easily be found through its equation.

The general form of a horizontally opening parabola is \( y^2 = 4p(x - h) \). In this scenario, the vertex is expressed by the point \((h, 0)\). For our specific parabola, the vertex appears at \((a^2, 0)\).

Vertices act as reference points to assist with finding intersections or calculating geometric parameters such as the area or focus. Understanding where a parabola's vertex lies gives insight into its central symmetry and helps address more complex mathematical questions efficiently. Recognizing that the vertex is a point on the x-axis that defines the symmetry of this parabola enriches our understanding of this mathematical shape.