Problem 33 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the line \(\frac{x}{a}+\frac{y}{b}=1\) moves in such a way that \(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}\), where \(c\) is a constant, then the locus of the foot of the perpendicular from the origin on the straight line describes the circle (A) \(x^{2}+y^{2}=4 c^{2}\) (B) \(x^{2}+y^{2}=2 c^{2}\) (C) \(x^{2}+y^{2}=c^{2}\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (C) \(x^2 + y^2 = c^2\).

Step 1: Equation of the Perpendicular

The equation of line is given as \(\frac{x}{a} + \frac{y}{b} = 1\). We first need to find the foot of the perpendicular from the origin to this line. The slope of the line is \(-\frac{a}{b}\), and the slope of the perpendicular line through the origin is \(\frac{b}{a}\).

Step 2: Foot of the Perpendicular Calculation

For the perpendicular from the origin to the line, the equation of the perpendicular line is \(y = \frac{b}{a}x\). Substituting this in the line equation \(\frac{x}{a} + \frac{y}{b} = 1\), we get: \[ \frac{x}{a} + \frac{\left(\frac{b}{a}x\right)}{b} = 1 \] Simplify to find that x-coordinate of the foot: \(x = \frac{a^{2}}{a^{2} + b^{2}} \). The y-coordinate can be found using \(y = \frac{b}{a}x = \frac{ab}{a^{2} + b^{2}} \).

Step 3: Locus Equation Derivation

The coordinates of the foot of the perpendicular are \(\left(\frac{a^{2}}{a^{2}+b^{2}}, \frac{ab}{a^{2}+b^{2}}\right)\). We look for the locus of this point. Using \(\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^2}\), we have \(\frac{a^{2}b^{2}}{a^{2}+b^{2}} = c^{2} \), which relates to the circle equation \(x^{2} + y^{2}\).

Step 4: Simplification and Circle Locus

Using the earlier expression \(\left(\frac{a^{2}}{a^{2}+b^{2}}, \frac{ab}{a^{2}+b^{2}}\right)\), and squaring these we have: \[ \left(\frac{a^{4}}{(a^{2}+b^{2})^{2}} + \frac{a^{2}b^{2}}{(a^{2}+b^{2})^{2}}\right) = \frac{a^{4} + a^{2}b^{2}}{(a^{2}+b^{2})^{2}} \] This simplifies to \(\frac{a^{2}(a^{2} + b^{2})}{(a^{2}+b^{2})^{2}} = \frac{a^{2}}{a^{2}+b^{2}}\)Given \(\frac{a^{2}b^{2}}{a^{2}+b^{2}} = c^{2}\), the sum could be simplified directly as \(x^{2} + y^{2} = c^2\).Therefore, the correct answer is \(x^{2} + y^{2} = c^{2}\).

Step 5: Conclusion

Thus, by identifying the conditions and deriving the appropriate locus equation, we see that the foot of the perpendicular from the origin to the given line traces a path in the shape of a circle. The correct option is (C) \(x^2 + y^2 = c^2\).

Key Concepts

Foot of the Perpendicular

In geometry, especially within coordinate geometry, finding the foot of the perpendicular is a common task. Imagine a point dropping a vertical line down to another straight line, leading to the shortest path possible. This concept becomes very useful when working with different geometric constructions and proves vital for numerous problem-solving scenarios in mathematics.
This point of intersection, or the foot of the perpendicular, is achieved by ensuring the new line formed is perpendicular. In our exercise, given the line equation \(\frac{x}{a} + \frac{y}{b} = 1\), the slope is \(-\frac{a}{b}\). Hence, for the line through the origin to be perpendicular, it must have the slope \(\frac{b}{a}\).

• Derivative Perpendicular: Line slopes are negative reciprocals.
• Equation Formation: Use the origin's coordinates.
• Intersection Determination: Solving both equations simultaneously.

The calculated coordinates, as shown in half our steps, \(\left(\frac{a^{2}}{a^{2} + b^{2}}, \frac{ab}{a^{2} + b^{2}}\right)\), are crucial for further analysis. They allow us to understand the movement and describe broader geometric paths.

Equation of a Circle

The equation of a circle is a central theme not only because of its simplicity but also due to its pivotal role in various fields such as physics, engineering, and computer graphics. The standard form \(x^2 + y^2 = r^2\) presents a circle centered at the origin with radius \(r\). For the equation \(\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}\) in our problem, the task was to relate any line movement to a possible circular locus.
A significant portion of solving these problems involves recognizing the transformation or path a point traces under specific conditions. As demonstrated, substituting known expressions and manipulating equations yields the desired form, ultimately verifying it matches the circle equation.

• Circle Recognition: Mapping equations like \(x^2 + y^2 = c^2\).
• Locus Analysis: Seeing paths or loci align with familiar shapes.
• Relevance Verification: Ensuring derived formats equal circle descriptions.

This equation elegance makes these queries not just an exercise in calculation but an exploration in recognizing patterns and underlying geometric truths.

Coordinate Geometry

Coordinate geometry represents a beautiful blend of algebra and geometry, leveraging coordinate points to explore geometric properties and establishing a deep connection between various geometric figures' characteristics and their algebraic representations.
In this exercise, understanding the relationship between line equations and point positions on curves becomes evident. By abstractly applying these concepts, like finding the foot of a perpendicular and discussing locus formation, transformations are deeply illustrated and calculated. Whether acting as a bridge in understanding the behavior of a circle through algebra or helping visualize motion paths for physical interpretation, coordinate geometry plays a pivotal role.

• Visualizing Slopes and Positions: Handling line slopes and perpendiculars.
• Formulating Geometric Paths: Relating movement to algebraic forms.
• Translation into Real-World Motion: Using paths and loci descriptions practically.

Appreciating these interactions heightens comprehension, allowing students to not only solve exercises but also interconnect mathematical ideas, fostering a holistic understanding of geometry and its application in the analytical realm.