Problem 90 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The orthogonal trajectory of the family of parabolas \(y^{2}=4 a x\) is (A) \(2 x^{2}+y^{2}=c\) (B) \(x^{2}+2 y^{2}=c\) (C) \(2 x^{2}-y^{2}=c\) (D) None of these

Step-by-Step Solution

Answer

The orthogonal trajectory is (A) \(2x^2 + y^2 = c\).

Step 1: Differentiate the given equation

The family of parabolas is given by the equation \( y^2 = 4ax \). To find the orthogonal trajectory, we first differentiate this with respect to \( x \). Using implicit differentiation, we get \( 2y \frac{dy}{dx} = 4a \). Solving for \( \frac{dy}{dx} \), we find \( \frac{dy}{dx} = \frac{2a}{y} \).

Step 2: Find the negative reciprocal of the slope

Orthogonal trajectories have slopes that are negative reciprocals of each other. Thus, the slope of the orthogonal trajectory is \( \frac{dy}{dx} = -\frac{y}{2a} \).

Step 3: Express the parameter 'a' in terms of 'x' and 'y'

Given \( y^2 = 4ax \), we rearrange to find \( a = \frac{y^2}{4x} \). Substituting this into the expression for the orthogonal slope gives \( \frac{dy}{dx} = -\frac{y}{2 \times \frac{y^2}{4x}} = -\frac{2x}{y} \).

Step 4: Solve the differential equation

Now, we solve the differential equation \( \frac{dy}{dx} = -\frac{2x}{y} \) by separating variables: \( y\, dy = -2x\, dx \). Integrating both sides, we get \( \frac{y^2}{2} = -x^2 + C \), or rearranging gives \( x^2 + \frac{y^2}{2} = C \).

Step 5: Simplify to match given choices

To match the form of one of the given choices, we multiply through by 2 to clear the fraction: \( 2x^2 + y^2 = 2C \). Letting \( c = 2C \), we obtain \( 2x^2 + y^2 = c \), which matches option (A).

Key Concepts

Implicit Differentiation

Implicit differentiation is a valuable technique when dealing with equations where the variables are intertwined, or not easily separated. In this context, instead of solving for one variable in terms of the other and then differentiating, you differentiate both sides of the equation directly with respect to a common variable, typically \( x \). This approach allows us to find the derivative of \( y \) with respect to \( x \), even when \( y \) is not isolated. In the original exercise, we differentiate the equation \( y^2 = 4ax \) with respect to \( x \). This requires us to apply the chain rule to the \( y^2 \) term, resulting in \( 2y \frac{dy}{dx} = 4a \). Implicit differentiation is indispensable when seeking orthogonal trajectories where you need to find a perpendicular direction to a given family of curves.

Parabolas

Parabolas are conic sections defined as the locus of points equidistant from a fixed point known as the focus and a line called the directrix. The standard form of a parabola that opens horizontally, like the one in the exercise, is given by \( y^2 = 4ax \), where \( a \) represents the distance from the vertex to the focus.
In solving the orthogonal trajectories, we start with this family of parabolas and understand their geometric properties, such as symmetry and vertex location. Understanding how these parabolas are structured is crucial to determining the slopes of lines perpendicular to them, which leads to finding their orthogonal trajectories. Recognizing that the family of parabolas share the same form allows us to manipulate the equations and derive other useful properties.

Differential Equation

Differential equations are equations that involve derivatives, expressing rates of change. They are essential in modeling phenomena where change is dynamic. In this exercise, once the slope \( \frac{dy}{dx} = -\frac{2x}{y} \) is found, an equation that describes the rate of change for the orthogonal trajectory is established.
To solve it, we use a technique called separation of variables. This involves rearranging the equation into forms where each side includes only one of the variables, and then integrating both sides. So, here it transforms into \( y \, dy = -2x \, dx \).
Upon integration, the solution \( \frac{y^2}{2} = -x^2 + C \) is found. Simplifying and matching the form to one of the provided options gives the orthogonal trajectory equation. Differential equations are the mathematical backbone in translating slope relationships into comprehensible and precise trajectory curves.