Problem 80 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The locus of the vertices of the family of parabolas \(y=\frac{a^{3} x^{2}}{3}+\frac{a^{2} x}{2}-2 a\) is (A) \(x y=\frac{105}{64}\) (B) \(x y=\frac{3}{4}\) (C) \(x y=\frac{35}{16}\) (D) \(x y=\frac{64}{105}\)

Step-by-Step Solution

Answer

The correct choice is (B) \(x y = \frac{3}{4}\).

Step 1: Identify the Vertex of the Parabola

The vertex form of a parabola is typically easy to identify in the form \( y = Ax^2 + Bx + C \). To find the vertex, we use the formula for the x-coordinate of the vertex \( x = -\frac{B}{2A} \). Here, \( A = \frac{a^3}{3} \) and \( B = \frac{a^2}{2} \).

Step 2: Calculate x-coordinate of the Vertex

Using the vertex formula, the x-coordinate becomes \( x = -\frac{\frac{a^2}{2}}{2\cdot\frac{a^3}{3}} = -\frac{3}{4a} \).

Step 3: Substitute x back to find y-coordinate

Substitute \( x = -\frac{3}{4a} \) back into the equation to find the y-coordinate, which gives: \[ y = \frac{a^3}{3}\left(-\frac{3}{4a}\right)^2 + \frac{a^2}{2}\left(-\frac{3}{4a}\right) - 2a \].

Step 4: Simplify the y-coordinate expression

Calculate each term: \Let: \( y = \frac{a^3}{3}\cdot\frac{9}{16a^2} - \frac{3a^2}{8a} - 2a \) This simplifies to \( \frac{3a}{16} - \frac{3}{8} - 2a \).

Step 5: Combine all terms in the y-expression

Simplify further: \\( \frac{3a}{16} - \frac{3}{8} - 2a = \frac{3a - 6 - 32a}{16} = \frac{-29a - 6}{16} \).

Step 6: Derive the result for the locus equation

The vertex \( (x, y) \) is \( \left(-\frac{3}{4a}, \frac{-29a - 6}{16}\right) \). Find the product: \( xy = \left(-\frac{3}{4a}\right) \left(\frac{-29a - 6}{16}\right)\).

Step 7: Simplify the locus equation

Upon simplification, \\( xy = \frac{3}{4a} \cdot \frac{29a + 6}{16} \). As \( a \) cancels, \( xy = \frac{3}{4} \), which matches option B.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a key feature in understanding its shape and position on the coordinate plane. It is the point where the parabola changes direction, either turning upward or downward. The vertex can be thought of as the 'peak' or 'lowest point' depending on the orientation of the parabola.
For quadratic equations in standard form, which is given by \(y = Ax^2 + Bx + C\), the x-coordinate of the vertex can be calculated using the formula \(x = -\frac{B}{2A}\).

• The vertex represents the maximum or minimum value of the quadratic function.
• It is crucial for graphing as it helps define the parabola's axis of symmetry.
• This point provides the optimal solution in many real-world applications.

In the exercise, using the formula for the vertex, \(x = -\frac{B}{2A}\), the x-coordinate of the vertex for the given parabola is found to be \(-\frac{3}{4a}\).
This calculation helps establish the formula or the locus of the vertices of the family of parabolas as examined in the problem.

Quadratic Equation

A quadratic equation is a foundational concept in algebra and coordinate geometry. It is typically expressed in the form \(y = Ax^2 + Bx + C\). This equation represents a parabola when graphed on the coordinate plane.

• Quadratics are second-degree polynomial equations; the highest power of the variable is 2.
• The number of real solutions of a quadratic equation depends on the discriminant \(B^2 - 4AC\).
• It can be solved using several methods including factoring, completing the square, and the quadratic formula.

The general shape of a quadratic graph is a parabola. Depending on the leading coefficient \(A\):
- If \(A > 0\), the parabola opens upwards, resembling a U-shape.
- If \(A < 0\), the parabola opens downwards, resembling an inverted U-shape.

In coordinate geometry, identifying and extracting the variables \(A\), \(B\), and \(C\) from a quadratic equation is essential for solving problems related to parabolas, as seen in the given exercise.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometric figures through the use of a coordinate system. This makes it possible to geometrically interpret algebraic expressions and equations.
It is a powerful tool that lays the foundation for linking algebra and geometry. It makes it easier to visualize mathematical concepts on a graph.

• Coordinate geometry uses a pair of numbers to define the position of points on the plane.
• It helps in solving various geometry problems using algebraic techniques and vice versa.
• Understanding the features of lines, circles, and parabolas becomes feasible.

For instance, in this exercise, coordinate geometry helps in determining the locus. Locus refers to the set of all points, typically forming a geometric figure, that satisfies given conditions.
By using coordinate geometry, we can identify the behavior and position of the set of points which forms the locus for the vertices of the family of parabolas.