Problem 29 - A Complete Resource Book in Mathematics for JEE Main

Question

The differential equation of the curve for which the normal at every point passes through a fixed point \((h, k)\) is (A) \(y-k=\frac{d x}{d y}(h-x)\) (B) \(y-k=\frac{d x}{d y}(x-h)\) (C) \(y-k=\frac{d y}{d x}(h-x)\) (D) \(y-k=\frac{d y}{d x}(x-h)\)

Step-by-Step Solution

Answer

(C) \(y-k=\frac{d y}{d x}(h-x)\) is the correct choice.

Step 1: Understand the Problem

We need to find the differential equation that describes a curve where the normal at every point \(x, y\) passes through a fixed point \(h, k\). The normal is perpendicular to the tangent at a point on a curve.

Step 2: Relate the Normal to the Tangent

For a curve, if the slope of the tangent at a point \(x, y\) is given by \(\frac{dy}{dx}\), then the slope of the normal at that point is \(-\frac{dx}{dy}\). This is because the normal is perpendicular to the tangent.

Step 3: Equation of the Normal

For a line with a known slope, the normal line passing through a point \(x, y\) is given by \(y - y_1 = m(x - x_1)\). Here, \(-\frac{dx}{dy}\) is the slope of the normal line. So, substituting, the normal through the fixed point \(h, k\) becomes: \(y - k = -\frac{dx}{dy}(x - h)\).

Step 4: Convert Equation Form

Rearrange \(y - k = -\frac{dx}{dy}(x - h)\) to get it in a recognizable form by changing the negative sign present: \(y - k = \frac{dy}{dx}(h - x)\). This matches the structure of one of the given options.

Step 5: Confirm the Correct Option

Compare the derived equation \(y - k = \frac{dy}{dx}(h - x)\) to the options given and we find it matches Option C.

Key Concepts

Normal Line to a Curve

Finding a normal line to a curve is a crucial aspect in differential calculus. A normal line is perpendicular to the tangent at a specific point on the curve.
This means that while the tangent represents the slope at a point, the normal deviates at the steepest possible angle.
Understanding the relationship between these two can help us analyze how curves behave in different scenarios.

• Normal lines are relevant in geometry, especially when determining the shortest path or distance related to curves.
• They provide insight into the geometrical structure and orientation of a curve in two-dimensional space.

Once you grasp the concept of normal lines, linking them to tangent lines becomes much easier.

Slope of Tangent and Normal

The slope of a tangent line at a point on a curve is a primary factor that describes the rate of change of the curve at that point.
For instance, if you have the derivative \( \frac{dy}{dx} \), it tells you the slope of the tangent at that point.
Understanding how tangents and normals relate helps visualize how curves interact with straight lines:

• The slope of a tangent line, as described by \( \frac{dy}{dx} \), represents the immediate rate of change on the curve.
• The normal line, being perpendicular to the tangent, has a slope described by the negative reciprocal, \( -\frac{dx}{dy} \).

This inverse relationship indicates how changes in the curve's inclination affect its perpendicular counterpart.

Equation of a Line

To find the equation of a line, such as a normal line to a curve, one starts with a known point and a slope.
Using the point-slope form \( y - y_1 = m(x - x_1) \), you can plug in values to find specific line equations.
The resulting equation describes how steeply the line inclines or reclines when passing through specific points.

• Point-slope form is versatile in forming equations for lines given limited parameters.
• The slope \( m \) for normal lines connecting to fixed points helps in cross-verifying positioning related to curves.

Identifying these elements simplifies the task of illustrating the line's path and orientation.

Curve Characteristics

A curve has unique characteristics based on its equation and derivative properties.
When examining a curve where normal lines intersect a fixed point \((h, k)\), one uncovers distinct geometric and algebraic properties.
Approaching these problems often involves:

• Using derivatives like \( \frac{dy}{dx} \) to discern the slope.
• Applying these to understand how normal lines originate and extend from points on the curve.
• Identifying how these interconnections define the curve's broader geometric framework.

In these analyses, the differential equation of curves captures dynamic changes and configuration in precise mathematical terms.