Problem 26 - A Complete Resource Book in Mathematics for JEE Main 2017

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

n is equal to 2.

Step 1: Understand the General Form of the Curve

The equation given is \(x^n y = a^n\). This implies we are studying a curve and its properties associated with tangents that make a triangle with the axes.

Step 2: Derive the Slope of the Tangent

Consider a point \((x_1, y_1)\) on the curve. Differentiate \(x^n y = a^n\) implicitly with respect to \(x\): \[\frac{d}{dx}(x^n y) = \frac{d}{dx}(a^n)\] which gives \[nx^{n-1} y + x^n \frac{dy}{dx} = 0\]. Solving for \(\frac{dy}{dx}\), we get \[\frac{dy}{dx} = -\frac{n y}{x}\].

Step 3: Equation of the Tangent Line

Using point-slope form, the equation of the tangent line at \((x_1, y_1)\) is \[y - y_1 = -\frac{n y_1}{x_1}(x - x_1)\]. Simplifying this, the tangent is \[nx_1y + xy_1 = 2a^n\].

Step 4: Find Intercepts with Axes

The intercepts with the axes can be found from the tangent equation. For the y-intercept, set \(x = 0\): \(nx_1y = 2a^n\Rightarrow y = \frac{2a^n}{nx_1}\). For the x-intercept, set \(y = 0\): \(xy_1 = 2a^n\Rightarrow x = \frac{2a^n}{y_1}\).

Step 5: Calculate the Area of the Triangle

The area \(A\) of the triangle formed by the tangent and the axes is \(A = \frac{1}{2} \times x\text{-intercept} \times y\text{-intercept}\). Substitute the values, \[A = \frac{1}{2} \times \frac{2a^n}{nx_1} \times \frac{2a^n}{y_1} = \frac{2a^{2n}}{nx_1y_1}\].

Step 6: Solve for n with Constant Area

Since the area is constant, \(\frac{2a^{2n}}{nx_1y_1}\) is constant. For it to be constant as independent of \(x_1, y_1\), these terms should cancel, i.e., each term's exponents of \(x\) and \(y\) should equal the degree of a constant term. Matching powers, \(n = 2\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus when you have functions defined by equations that cannot be easily solved for one variable in terms of another. Some equations involve variables that are intertwined in such a way that isolating one might prove difficult or impossible. In such cases, implicit differentiation allows us to find the derivative without needing to explicitly solve for one of the variables in terms of the other.

Here's a simple way to approach it:

• Differentiate both sides of the equation with respect to the same variable, usually "x".
• Whenever you differentiate a term involving "y", treat "y" as a function of "x", which means using the chain rule you'll multiply by \( \frac{dy}{dx} \) to represent the derivative of "y".
• Simplify the resulting equation to solve for \( \frac{dy}{dx} \).

In the context of the given exercise, implicit differentiation was used to find the slope of the tangent line to the curve given by the equation \(x^n y = a^n\). Since explicit solutions might be complex, implicit differentiation is a powerful tool to find derivatives and, subsequently, slopes of tangent lines for such curves.

Area of a Triangle

The area of a triangle is a fundamental concept in geometry with diverse applications in mathematics. For a triangle on the coordinate plane, especially one formed by the intercepts, calculating the area can involve the vertices' locations.

For a right triangle with legs on the coordinate axes and formed by a tangent line:

• The x-intercept and y-intercept provide the length of the base and height, respectively.
• The area \(A\) of the triangle can be calculated using the formula \(A = \frac{1}{2} \times \text{base} \times \text{height}\).

In our exercise, we considered the intercepts given for the tangent to the curve \(x^{n} y=a^{n}\). We used the formula with the intercepts determined from the tangent line equation to find that \( \frac{2a^{2n}}{nx_1y_1} = \text{constant} \). This setup helped in determining that \(n = 2\) is the value making the area constant, satisfying the problem's condition.

Intercepts of Tangents

Intercepts in the context of tangent lines refer to the points where the tangent intersects the axes. These points tell us about the spatial relationship of the curve with respect to the coordinate axes and are essential in solving different mathematical problems involving curves and their tangent lines.

Here's a basic understanding:

• The x-intercept is found by setting \(y = 0\) in the tangent's equation and solving for \(x\).
• The y-intercept is obtained by setting \(x = 0\) and solving for \(y\).

In our exercise, for the curve \(x^n y = a^n\), differentiating the equation and forming the tangent allowed us to derive:

• y-intercept as \( \frac{2a^n}{nx_1} \)
• x-intercept as \( \frac{2a^n}{y_1} \)

Using these values to form a triangle and calculate its area led us to understanding the constant nature of the area and establishing \(n = 2\), reflecting the balance between the geometry of tangents and their intercepts.