Problem 26 - A Complete Resource Book in Mathematics for JEE Main

Question

If the area of the triangle included between the axes and any tangent to the curve \(x^{n} y=a^{n}\) is constant, then \(n\) is equal to (A) 1 (B) 2 (C) \(\frac{3}{2}\) (D) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of \(n\) is 1.

Step 1: Understand the problem

We have a curve given by the equation \(x^n y = a^n\). We need to find the value of \(n\) such that the area of the triangle formed by any tangent to this curve with the coordinate axes is constant. This area would depend on the intersection points of the tangent line with the axes.

Step 2: Find the derivative of the curve

Given the equation of the curve as \(x^n y = a^n\), we differentiate implicitly to find the slope of the tangent \(m\). Differentiating with respect to \(x\), we have \(n x^{n-1} y + x^n \frac{dy}{dx} = 0\), from which we derive the differential \(\frac{dy}{dx} = -\frac{n x^{n-1} y}{x^n} = -\frac{n y}{x}\). So the slope of the tangent is \(m = -\frac{n y}{x}\).

Step 3: Equation of the tangent

The equation of the tangent line at a point \((x_1, y_1)\) on the curve is given by \(y - y_1 = -\frac{n y_1}{x_1}(x - x_1)\). This simplifies to \(y = -\frac{n y_1}{x_1}x + y_1 + \frac{n y_1^2}{x_1}\).

Step 4: Calculate the x-intercept and y-intercept of the tangent

The x-intercept occurs when \(y = 0\), giving \(0 = -\frac{n y_1}{x_1}x + y_1 + \frac{n y_1^2}{x_1}\), which leads to \(x = x_1 + n y_1\). The y-intercept occurs when \(x = 0\), giving \(y = y_1 + \frac{n y_1^2}{x_1}\).

Step 5: Determine the area of the triangle

The vertices of the triangle are \((0,0), (x_1 + n y_1, 0), (0, y_1 + \frac{n y_1^2}{x_1})\). The area \(A\) of the triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\), which becomes \(A = \frac{1}{2} \times (x_1 + n y_1) \times (y_1 + \frac{n y_1^2}{x_1})\).

Step 6: Simplify the expression for the area

Substitute the expression for \(y_1 = \frac{a^n}{x_1^n}\) (since \(y = \frac{a^n}{x^n}\) on the curve) into the area equation. After simplification, the area \(A\) becomes \(\frac{a^n(n+1)}{2 x_1^{n-1}}\).

Step 7: Determine the value of n

For the area \(A\) to be constant, the exponent of \(x_1\) in the simplified area expression must be zero. Thus, solving \(n-1=0\), we get \(n = 1\). Therefore, the value of \(n\) is equal to 1.

Key Concepts

Tangent to a Curve

Imagine a curve plotted on a graph. A tangent line is a straight line that touches the curve at just one point. It only kisses, not crosses, the curve at this point. This special line will have the exact same slope as the curve at that point.

To find the tangent to a curve given by an equation like \(x^n y = a^n\), we use a technique called implicit differentiation. This method helps us find the slope of the tangent even when \(y\) cannot be easily separated from \(x\).

The slope of the tangent is essential because it helps us write the equation of the tangent line. Once you have the slope, you can use a point on the curve (where the line touches it) to express the equation of the tangent. This is super important when solving problems like determining areas formed by tangents.

Triangle Area

The area of a triangle is calculated by the simple formula \(A = \frac{1}{2} \times \text{base} \times \text{height}\). When a tangent line forms a triangle with the coordinate axes, it has a vertex at the origin \((0, 0)\) and intersections with the x-axis and y-axis.

In our case, from the tangent line's equation, we find these intersections—known as intercepts. Let's say the x-intercept is \((x_1 + n y_1, 0)\) and the y-intercept is \((0, y_1 + \frac{n y_1^2}{x_1})\), these become the base and height of the triangle.

By substituting these intercepts back into the area formula, we get the triangle's area as \(A = \frac{1}{2} \times (x_1 + n y_1) \times (y_1 + \frac{n y_1^2}{x_1})\). If the area has to remain constant for every tangent, it leads to constraints on the curve's properties, like the value of \(n\).

Implicit Function Theorem

The implicit function theorem is a powerful mathematical tool that helps when functions are not given explicitly. Often, the relations between variables are tangled, like in the equation \(x^n y = a^n\).

Instead of solving for one variable in terms of the other outright, the implicit function theorem gives us a way to work with derivatives directly. We apply implicit differentiation to get the slope \(\frac{dy}{dx}\). This slope is crucial because it gives us the equation of the tangent—essentially unwrapping the curve's hidden geometry.

Using implicit differentiation, we found \(\frac{dy}{dx} = -\frac{n y}{x}\), which tells us how the dependent variable \(y\) changes with \(x\). It's like peeking behind the curtain at how each change in \(x\) reflects in \(y\), even when \(y\) isn't isolated. Implicit differentiation thus lets us handle complex curves with grace and ingenious strategy.