Problem 81 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The point of intersection of the lines \(\left(a^{3}+3\right) x+a y+a-3=0\) and \(\left(a^{5}+2\right) x+(a+2) y+2 a+3=0\) (a real) lies on the \(y\)-axis for |Online May 7, 2012] (a) no value of \(a\) (b) more than two values of \(a\) (c) exactly one value of \(a\) (d) exactly two values of \(a\)

Step-by-Step Solution

Answer

(d) exactly two values of \(a\)

Step 1: Identify Intersection Conditions

The point of intersection of two lines lies on the y-axis if its x-coordinate is zero. We will find the intersection of the given lines and ensure this condition is met.

Step 2: Eliminate x From Equations

Since the point lies on the y-axis, set the x-coordinate to zero in both equations: \( (a^3+3)(0) + ay + a - 3 = 0 \) which simplifies to \( ay + a - 3 = 0 \). Similarly, the second equation becomes \( (a^5+2)(0) + (a+2)y + 2a + 3 = 0 \) simplifying to \( (a+2)y + 2a + 3 = 0 \).

Step 3: Solve for y in Both Equations

From the first equation \( ay + a - 3 = 0 \), we solve for y to get \( y = \frac{3-a}{a} \). From the second equation \( (a+2)y + 2a + 3 = 0 \), we solve for y to get \( y = -\frac{2a+3}{a+2} \).

Step 4: Equate the Two y-values

Since both expressions represent the y-coordinate of the same point, set \( \frac{3-a}{a} = -\frac{2a+3}{a+2} \) and solve for \(a\).

Step 5: Cross-Multiply and Rearrange

Cross-multiply to obtain \( (3-a)(a+2) = -(2a+3)a \), and expand to get \( 3a + 6 - a^2 - 2a = -2a^2 - 3a \). Simplify to \( a^2 + 8a + 6 = 0 \).

Step 6: Solve Quadratic Equation

Use the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) on \( a^2 + 8a + 6 = 0 \) to find \( a = \frac{-8 \pm \sqrt{64 - 24}}{2} \). Calculate the results to find \( a = -4 \pm \sqrt{10} \).

Step 7: Assess the Number of Solutions

There are two distinct solutions: \( a = -4 + \sqrt{10} \) and \( a = -4 - \sqrt{10} \). So, there are exactly two values of \(a\).

Key Concepts

Coordinate Geometry

In coordinate geometry, the relationship between points, lines, and shapes on a plane is a central study point. One common task is finding the point of intersection between two lines. This point is where the two lines meet on the plane.

To find a point of intersection, you typically solve the equations of the two lines simultaneously, meaning you find a common solution for both equations. For lines in two-dimensional space, their equations are often linear, represented in the form \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.

• Lines intersect at a single point if they are not parallel.
• If they are the same line, there are infinitely many points of intersection.
• If they are parallel, they do not intersect.

In this problem, intersecting lines have been checked to ensure that their point of intersection lies on the y-axis. This occurs when the x-coordinate of the point of intersection is zero.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, usually in the form \(ax^2 + bx + c = 0\). Finding their roots is crucial, as it helps in locating specific values that satisfy certain conditions, such as intersections in coordinate geometry.

In the exercise, to solve for the parameter \(a\) where the intersection lies on the y-axis, a quadratic equation emerges, \(a^2 + 8a + 6 = 0\). Solving this equation employs the quadratic formula:

• The quadratic formula is \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• It delivers possible values for \(a\) based on any potential intersection.

This systematic approach provides the real values of \(a\) that satisfy the original line equations when intersected. It’s a critical step in ensuring that the conditions set by the geometry problem are met.

System of Linear Equations

A system of linear equations refers to two or more linear equations considered at once, where a solution satisfies all the equations simultaneously. Solving such a system is a staple in algebra and is applicable in numerous scenarios including finding intersections of lines.

In this exercise, the intersection of two lines is investigated by creating a system of equations that need to be solved simultaneously.

• Each line in the problem is represented by a linear equation.
• The solution, or intersection point, must satisfy both equations regarding how they relate to x and y on the coordinate plane.
• Setting the x-coordinate to zero helps resolve where the point lies solely on the y-axis.

Systematic steps are followed to equate the y-values derived from each linear equation, leading to a quadratic equation. Solving this confirms the number of valid \(a\)-values, providing complete solutions to the problem.