Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the \(x\)-intercept of the normal and passing through \((2,1)\) is (A) \(x^{2}+y^{2}-x=0\) (B) \(4 x^{2}+2 y^{2}-9 y=0\) (C) \(2 x^{2}+4 y^{2}-9 x=0\) (D) \(4 x^{2}+2 y^{2}-9 x=0\)

Step-by-Step Solution

Answer

Option (B) is correct: \(4x^2 + 2y^2 - 9y = 0\).

Step 1: Understand the Problem Statement

The problem describes a curve where the square of the ordinate (y) is twice the area of a rectangle formed by the abscissa (x) and the x-intercept of the normal. The normal is perpendicular to the tangent of the curve and passes through the point (2,1). We need to find the right equation from the given options.

Step 2: Express the Relationships Mathematically

The equation in words is that the square of the ordinate \( y \) is twice the rectangle formed by \( x \) (abscissa) and the x-intercept of the normal. Therefore, \( y^2 = 2x \times \text{x-intercept of the normal} \).

Step 3: Determine the x-intercept of the Normal

The x-intercept of the normal to the curve at a point \((x_1,y_1)\) is given by \( x_1 - \frac{y_1}{f'(x_1)} \), where \( f'(x) \) is the derivative of the function. We'll use this later when finding the specific curve equation.

Step 4: Use the Point (2,1) as a Given

Since the curve passes through point (2,1), substitute this into any derived relationships to find specific constants or requirements as needed. Ensure the equation satisfies this point.

Step 5: Review the Given Options

Quickly verify which of the given options fit the description by checking the point (2,1) within each and ensuring the ordinate condition holds true.

Key Concepts

Ordinate and Abscissa relationship

In geometry and algebra, understanding the relationship between abscissa and ordinate is crucial, especially when dealing with the equation of a curve. The abscissa is a fancy term for the x-coordinate, and the ordinate is another name for the y-coordinate. They help describe the position of a point in a two-dimensional plane.
For the problem at hand, the key relationship given is that the square of the ordinate (or y-value) of a point on the curve is twice the product of the abscissa (the x-value) and another number related to the curve's geometry. Specifically, in this case, it is the x-intercept of the normal line at that point. Represented mathematically, this relationship can be written as:

• The equation is: \[ y^2 = 2x \times \text{x-intercept of the normal} \]

This means for any point \(x, y\), if you take the y-value, square it, and then calculate twice the area of a rectangle formed by x and some constant number, the two should be equal. This relationship is essential for finding what kind of curve this equation forms.

Normal to a curve

When we talk about the normal to a curve, we are referring to a specific line that is perpendicular to the tangent at any given point on the curve. The normal plays a significant role in the formulation of the curve's equation, particularly when dealing with problems related to geometric applications.


The normal line can be visualized as a line that 'hits' the curve exactly at a small, localized spot and stretches out perpendicularly. The x-intercept of this normal line, which we are interested in, is derived from the point of tangency. At any point \(x_1, y_1\) on the curve, if we know the slope of the tangent is given by the derivative \(f'(x_1)\), the x-intercept is calculated as follows:

• The x-intercept of the normal:\[ x_1 - \frac{y_1}{f'(x_1)} \]

The subtraction here accounts for the fact that the normal line stretches away from the point of tangency into the universe of x-values until it hits the x-axis (that's where it intercepts the axis). Finding this intercept is key to establishing the relationship described in the problem.

Derivative and tangent line

The concept of the derivative is central when we start discussing tangent lines to a curve. In essence, a derivative gives us the slope of the tangent line at any given point on the curve. This tangent line is imagined as a line that just "touches" the curve at one precise point—much like touching a ball with a finger at its crest, without drilling into it.

Calculating the derivative of a function is pivotal since it tells us the exact angle at which the curve is inclined at that point, and hence the exact slope of the tangent line. Once the derivative \(f'(x_1)\) is determined, this slope becomes:

• The slope of the tangent line at \(x_1\) is \(f'(x_1)\)

With the slope known, you can formulate the equation of the tangent line using the point-slope formula:

• Equation of the tangent line:\[ y - y_1 = f'(x_1)(x - x_1) \]

This equation provides a complete description of the tangent line at that specific point, illustrating how it lightly trails along the curve without ever intersecting it dramatically. Understanding both the tangent and normal lines are essential for tackling curve equations efficiently.