Problem 28 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A light ray emerging from the point source placed at \(P(2,3)\) is reflected at point ' \(\theta\) on the \(y\)-axis and then passes through the point \(R(5,10)\). Coordinates of ' \(Q\) ' are (A) \((0,3)\) (B) \((0,2)\) (C) \((0,5)\) (D) none of these

Step-by-Step Solution

Answer

The coordinates of point Q are (0, 5).

Step 1: Determine the Equation of Line from P to Q

The point source is at \(P(2,3)\) and point \(Q\) is on the \(y\)-axis, so \(Q\) is \((0, y_q)\). The slope \(m_1\) of line segment \(PQ\) is found by:\[ m_1 = \frac{y_q - 3}{0 - 2} = \frac{y_q - 3}{-2} \]

Step 2: Determine the Equation of Line from Q to R

Since \(Q\) is \((0, y_q)\) and \(R\) is \((5, 10)\), the slope \(m_2\) of line segment \(QR\) is:\[ m_2 = \frac{10 - y_q}{5 - 0} = \frac{10 - y_q}{5} \]

Step 3: Reflective Property of the Slopes

For reflection, the slope \(m_1\) and \(m_2\) are related by \(m_1 \cdot m_2 = -1\). Substituting the expressions for \(m_1\) and \(m_2\):\[ \left( \frac{y_q - 3}{-2} \right) \cdot \left( \frac{10 - y_q}{5} \right) = -1 \]

Step 4: Solve for \(y_q\) using Reflective Equation

We solve the equation:\[ \frac{(y_q - 3)(10 - y_q)}{-10} = -1 \]Multiplying throughout by \(-10\),\[ (y_q - 3)(10 - y_q) = 10 \]\[ 10y_q - y_q^2 - 30 + 3y_q = 10 \]Combine like terms:\[ 13y_q - y_q^2 = 40 \]Rearrange to:\[ y_q^2 - 13y_q + 40 = 0 \]

Step 5: Solve the Quadratic Equation

To solve \(y_q^2 - 13y_q + 40 = 0\), use the quadratic formula, \(y_q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -13\), and \(c = 40\):\[ y_q = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 1 \cdot 40}}{2} \]\[ y_q = \frac{13 \pm \sqrt{169 - 160}}{2} \]\[ y_q = \frac{13 \pm 3}{2} \]This gives \(y_q = 8\) or \(y_q = 5\).

Step 6: Conclusion

Since \(Q\) is on the \(y\)-axis and must be a reflection point, evaluate both possibilities for logical consistency, and determine \(Q(0, 5)\) meets all conditions required by the geometry involved.

Key Concepts

Reflection of Light

When a light ray hits a surface and bounces off, it's known as the reflection of light. This concept is crucial in coordinate geometry for solving problems like the one given here. The point sources of light, like our given point at \((2,3)\), emit rays that follow a predictable path when reflected. Understanding this behavior helps us pinpoint where the ray will end up.

• The angle at which the light hits a surface is equal to the angle at which it reflects off.
• In coordinate geometry, the line representing the light's path can be defined using equations that describe its slope and position.

In our exercise, determining the reflection involves calculating where the line would intersect the \(y\)-axis based on the given conditions. The midpoint of this reflection, point \(Q\), is calculated by looking at the coordinates derived from the slopes of the light's path pre- and post-reflection.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation and takes the standard form of \(ax^2 + bx + c = 0\). It's fundamental for solving problems involving parabolas and dynamics that result in squared variables.In our task, we derived the quadratic equation from the relationship of the slopes, where we ultimately needed to find the \(y\)-coordinate of point \(Q\) on the \(y\)-axis. With the equation \( y_q^2 - 13y_q + 40 = 0\), solving it gives us the potential points of reflection.

• By using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we can find the values of \(y\) that satisfy our condition.
• In this case, we computed \(y_q = 8\) and \(y_q = 5\).

Finding the correct \(y_q\) involves verifying which output logically satisfies the conditions set by the light's path.

Slope of a Line

The slope of a line in a coordinate plane is a measure of its steepness and is a critical component of reflection problems. It is calculated as the ratio of the vertical change to the horizontal change between two points, often referred to as 'rise over run'.

• The slope \(m\) of a line through points \((x_1, y_1)\) and \((x_2, y_2)\) is determined by \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
• In our solution, the slopes \(m_1\) and \(m_2\) were crucial as they dictated the direction of the light's path before and after reflection.
• The property that \(m_1 \cdot m_2 = -1\), indicating perpendicular lines, was used to solve for \(y_q\).

By understanding how to calculate and use slopes, we can accurately determine how light reflects and trace its path through coordinate space.