Problem 133 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The points on the curve \(a y^{2}=x^{3}\) where the normal line makes equal intercepts on the axes are (A) \(\left(\frac{2 a}{9}, \frac{8 a}{27}\right)\) (B) \(\left(\frac{4 a}{9}, \frac{8 a}{27}\right)\) (C) \(\left(\frac{4 a}{9}, \frac{-8 a}{27}\right)\) (D) \(\left(\frac{4 a}{9}, \frac{4 a}{27}\right)\)

Step-by-Step Solution

Answer

The points are \( \left( \frac{4a}{9}, \frac{4a}{27} \right) \). Option D is correct.

Step 1: Given Equation and Condition

We start with the curve equation given by \( ay^2=x^3 \). The condition is that the normal makes equal intercepts on the axes. For a normal line with intercepts at \( c \), the equation is \( x + y = c \) (since both intercepts are equal \( c \) and negative slope indicates it should be \( x + y = c \)).

Step 2: Find the Derivative of the Curve

Find \( \frac{dy}{dx} \) for the given curve: Differentiate \( ay^2 = x^3 \) implicitly with respect to \( x \): \( \frac{d}{dx}(a y^2) = \frac{d}{dx}(x^3) \) leading to \( 2a y \frac{dy}{dx} = 3x^2 \). Thus, we have \( \frac{dy}{dx} = \frac{3x^2}{2ay} \).

Step 3: Calculate Slope of the Normal

The slope of the tangent \( m_t = \frac{3x^2}{2ay} \) and the slope of the normal \( m_n = -\frac{1}{m_t} = -\frac{2ay}{3x^2} \). Using the form of the line \( x+y=c \), the slope \( -1 \) confirms equal intercepts.

Step 4: Set the Slope of Normal to -1

Set \( -\frac{2ay}{3x^2} = -1 \). Solving yields:\[ \frac{2ay}{3x^2} = 1 \rightarrow 2ay = 3x^2 \].

Step 5: Substitute into Original Equation

Since \( 2ay = 3x^2 \), substitute \( y = \frac{3x^2}{2a} \) into \( ay^2 = x^3 \):\[ a\left(\frac{3x^2}{2a}\right)^2 = x^3 \]which simplifies to \[ \frac{9x^4}{4a} = x^3 \].

Step 6: Solve for x

Solve \( \frac{9x^4}{4a} = x^3 \) for \( x \):\[ 9x = 4a \rightarrow x = \frac{4a}{9} \].

Step 7: Solve for y

Using \( x = \frac{4a}{9} \), substitute back into \( y = \frac{3x^2}{2a} \):\( y = \frac{3(\frac{4a}{9})^2}{2a} \) leads to \( y = \frac{4a}{27} \).

Step 8: Check Solutions in Options

The point where the curve satisfies the condition is \( \left( \frac{4a}{9}, \frac{4a}{27} \right) \). This matches option D.

Key Concepts

calculus

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It is largely divided into differential calculus and integral calculus. Differential calculus is concerned with the concept of a derivative, which essentially measures how a function changes as its input changes. In this problem, we are interested in finding the derivative of the curve given by the equation \( ay^2 = x^3 \). Finding the derivative involves understanding the relationship between the two variables, \( x \) and \( y \), and how a small change in one affects the other.

To find the derivative \( \frac{dy}{dx} \), we are looking for the rate of change of \( y \) with respect to \( x \). This requires taking the implicit derivative of the given equation. By understanding derivatives, students can determine various properties about functions like their slopes, maxima, and minima. In this exercise specifically, we use calculus to find out where the normal line to a given curve is perpendicular to the line with equal intercepts on the axes.

coordinate geometry

Coordinate geometry, also known as analytic geometry, involves the study of geometric figures using a coordinate system. Here, we deal with curves, lines, and other geometric shapes represented on the Cartesian plane, which is defined by the x-axis and y-axis. In this particular problem, the curve defined by \( ay^2 = x^3 \) represents a set of points in the plane.

Understanding the curve on the plane helps to visualize where the normal to the curve makes equal intercepts on the axes, which means the normal line will intersect the x-axis at some point \( (c, 0) \) and the y-axis at \( (0, c) \). Both intercepts having the same value \( c \) implies the equation of the normal is \( x + y = c \).

By exploring relationships between points and slopes on these geometric objects on a coordinate plane, students can solve complex geometry problems involving lines, angles, and distances with a clear and structured approach.

implicit differentiation

Implicit differentiation is a technique used to find the derivative of functions that are not explicitly defined. In mathematics, many relationships between variables are given implicitly, meaning they are not solved for one variable in terms of another; instead, they are given in a form like \( ay^2 = x^3 \), which intertwines \( x \) and \( y \) without making \( y \) a function of \( x \) directly.

To apply implicit differentiation, you differentiate each side of the equation with respect to \( x \), treating \( y \) as a function of \( x \) (i.e., \( y = f(x) \)), which means you'll likely need to use the chain rule. For our example, differentiating gives \( 2a y \frac{dy}{dx} = 3x^2 \), and solving this yields the expression for \( \frac{dy}{dx} \).

This technique is incredibly useful for dealing with more complex relationships in advanced mathematics courses such as calculus, especially when it's challenging to separate the variables and solve for one in terms of another. By mastering implicit differentiation, students can analyze and understand curves that are more intricate than simple functions.