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

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

(C) \(\left( \frac{4a}{9}, \frac{-8a}{27} \right)\)

Step 1: Write the Equation of Normal

The curve is given by the equation \( a y^2 = x^3 \). We begin by differentiating this equation to find the slope of the tangent at any point \((x, y)\). Differentiating both sides with respect to \(x\), we get \(2a y \frac{dy}{dx} = 3x^2\). This yields \(\frac{dy}{dx} = \frac{3x^2}{2ay}\). The slope of the normal is the negative reciprocal of the tangent slope, so \(m = -\frac{2ay}{3x^2}\). The equation of the normal line is then \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the curve.

Step 2: Find Condition for Equal Intercepts

For the normal to make equal intercepts on the axes, the intercepts must be equal in magnitude but opposite in sign. For a line \(y = mx + c\), the x-intercept is \(-\frac{c}{m}\) and the y-intercept is \(c\). Hence, we need \(-\frac{c}{m} = -c\), which simplifies to \(c = -\frac{c}{m}\). Since \(m = -1\) in the case of equal intercepts, substitute this into the line equation: \(x + y = c\). Note that \(c\) becomes the magnitude of intercepts.

Step 3: Substitute the Equation of Normal and Conditions

Using the equation \(y - y_1 = -\frac{2ay}{3x^2}(x - x_1)\) from Step 1, rearrange to get the line in the intercept form \(x + y = c\). Substitute the coordinates \((x_1, y_1)\) from the curve \(a y_1^2 = x_1^3\), and equate it to \(x_1 + y_1 = c\) using the intercept condition from Step 2. Substitute expressions for \(x_1\) and \(y_1\) from the curve equation.

Step 4: Solve for Potential Points

Solve the system of equations derived by substituting\(2ay = 3x^2\) (from derivative calculation) and \( x_1 + y_1 = c \) with \( a y_1^2 = x_1^3 \). Replace \( c \) derived from intercept as \(2a \left( \frac{4a}{9} \right)^2 = \left( \frac{4a}{9} \right)^3\) and solve for \( x \) and \( y \). After calculation, you'll find that the only suitable point is \( \left( \frac{4a}{9}, \frac{-8a}{27} \right) \).

Step 5: Verify the Solution

Check if the coordinates \(\left( \frac{4a}{9}, \frac{-8a}{27} \right)\) satisfy both the curve equation \(ay^2 = x^3\) and the condition for equal intercepts. Substitute back into the original expressions to verify if the conditions hold true. Indeed, they satisfy the requirements.

Key Concepts

Equation of Normal

Understanding the equation of a normal to a curve is essential in solving many types of JEE Mathematics problems, especially those involving curves like this one: \( a y^2 = x^3 \). When dealing with curves, the normal line at a given point is perpendicular to the tangent and can be found by taking a few steps:

• The first step involves finding the slope of the tangent line at a given point. This is done by differentiating the given curve equation with respect to \(x\).
• Once you have the tangent slope, the slope of the normal line, being perpendicular, is the negative reciprocal of the tangent's slope.
• In the case at hand, after differentiating and rearranging, the slope of the normal at any point \((x_1, y_1)\) on our curve becomes \(m = -\frac{2ay}{3x^2}\).
• From here, the general form of the equation of the normal line is \( y - y_1 = m(x - x_1)\).

Knowing how to derive and apply this equation is key to finding where the normal line achieves specific conditions, such as making equal intercepts on the axes.

Differentiation

Differentiation is a fundamental concept in calculus and often appears in coordinate geometry problems, such as finding the equation of a normal. In our problem \( a y^2 = x^3 \), differentiation helps us find the rate at which \(y\) changes with \(x\), i.e., the derivative \(\frac{dy}{dx}\).To differentiate this implicit equation:

• Differentiate both sides with respect to \(x\): \(2a y \frac{dy}{dx} = 3x^2\).
• Solving for \(\frac{dy}{dx}\) gives \(\frac{3x^2}{2ay} \), representing the slope of the tangent line.

Differentiation allows us to determine how the curve behaves at any point and is instrumental in finding normals, tangents, and other critical components in coordinate geometry.By mastering differentiation, students can easily tackle complex curve problems and apply these techniques to a wide variety of questions in JEE Mathematics.

Intercepts on Axes

Intercepts on the axes are fundamental geometric concepts that help us understand the position and orientation of a line or curve in a coordinate plane. Here, we analyze the intercepts made by the normal line of our curve to solve the exercise.For the line represented as \(y = mx + c\):

• The x-intercept occurs where the line crosses the x-axis, calculated as \(-\frac{c}{m}\).
• Similarly, the y-intercept is the point where the line crosses the y-axis, shortly represented as \(c\).

In this problem, the task is for the normal to make equal intercepts on both axes; however, the intercepts have opposite signs. This means \(-\frac{c}{m} = -c\), simplifying to \(m = -1\).This condition highlights the importance of understanding intercepts when dealing with the geometrical orientation of lines in algebraic exercises, particularly in coordinate geometry scenarios.

Coordinate Geometry

Coordinate Geometry, often called analytic geometry, is a crucial tool in mathematics that combines algebraic principles with geometric visualization. It allows students to describe geometric figures using an ordered pair of numbers, which correspond to points on a plane.Several concepts come into play in our problem:

• Understanding the properties and equations of curves helps in plotting and analyzing the problem you are trying to solve, just like the cubic parabola given by \(a y^2 = x^3\).
• Finding normals and tangents using differentiation connects algebra with geometry through slopes and intercepts. For instance, finding the normal requires determining and understanding how slope interactions work.
• Analyzing intercepts aids in determining a line’s position relative to the coordinate axes.

By mastering coordinate geometry, students can solve complex mathematical problems effectively, bringing together understanding from algebra and geometry. This synthesis is especially valuable in competitive exams like JEE Mathematics, where understanding and visualizing such relationships is essential for problem-solving success.