Problem 22 - A Complete Resource Book in Mathematics for JEE Main

Question

Tangent is drawn to the ellipse \(\frac{x^{2}}{27}+y^{2}=1\) at \((3 \sqrt{3} \cos \theta, \sin \theta)\), where \(\theta \in(0, \theta / 2)\). Then, the value of \(\theta\) such that sum of intercepts on axes made by this tangent is minimum, is (A) \(\frac{\pi}{3}\) (B) \(\frac{\pi}{6}\) (C) \(\frac{\pi}{8}\) (D) \(\frac{\pi}{4}\)

Step-by-Step Solution

Answer

The value of \(\theta\) is \(\frac{\pi}{6}\).

Step 1: Identify the standard equation of an ellipse

The given ellipse is \(\frac{x^2}{27} + y^2 = 1\). This is in the standard form where \(a^2 = 27\) and \(b^2 = 1\), making \(a = \sqrt{27}\) and \(b = 1\).

Step 2: Write the tangent equation at a point on the ellipse

The general tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at a point \((x_0, y_0)\) is \(\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1\). Substituting \(a = \sqrt{27}\), \(b = 1\), and the given point \((3 \sqrt{3} \cos \theta, \sin \theta)\), the tangent equation becomes \(\frac{3\sqrt{3} \cos\theta \cdot x}{27} + \sin\theta \cdot y = 1\). Simplifying, we have \(\frac{x \cos\theta}{3\sqrt{3}} + y \sin\theta = 1\).

Step 3: Find x-intercept and y-intercept of the tangent

To find the x-intercept, set \(y = 0\) in the tangent equation \(\frac{x \cos\theta}{3\sqrt{3}} + y \sin\theta = 1\), which gives \(x = 3\sqrt{3} \sec\theta\). For the y-intercept, set \(x = 0\), giving \(y = \csc\theta\). So, the intercepts are \(3\sqrt{3} \sec\theta\) and \(\csc\theta\).

Step 4: Set up the expression for the sum of intercepts

The sum of intercepts on the axes made by the tangent is given by \(S(\theta) = 3\sqrt{3} \sec\theta + \csc\theta\). Our goal is to minimize this expression.

Step 5: Differentiate S(θ) and find critical points

To find the minimum, differentiate \(S(\theta)\) with respect to \(\theta\): \(\frac{d}{d\theta} S(\theta) = \frac{d}{d\theta}(3\sqrt{3} \sec\theta + \csc\theta) = 3\sqrt{3} \cdot \sec\theta \tan\theta + \cdot (-\csc\theta \cot\theta).\)Set this derivative to zero and solve for \(\theta\). This yields the critical points that may provide the minimum value of \(S(\theta)\).

Step 6: Solve the derivative equation

Solving the derivative equation yields \(3\sqrt{3} \sec\theta \tan\theta - \csc\theta \cot\theta = 0.\)Rearrange to find \(3\sqrt{3}\sin\theta = \cos^2\theta.\)Using trigonometric identities gives \(\theta = \frac{\pi}{6}\).

Step 7: Verify the result and determine the minimum intercept

Substitute \(\theta = \frac{\pi}{6}\) back into \(S(\theta) = 3\sqrt{3} \sec\theta + \csc\theta\) to verify that this value indeed provides a minimum. With \(\theta = \frac{\pi}{6}\), the sum of intercepts is minimized.

Key Concepts

Tangent to Ellipse

When dealing with ellipses, tangents can be quite fascinating. A tangent is a straight line that just "touches" an ellipse at exactly one point. The concept of drawing a tangent to an ellipse involves using the standard formula for the ellipse and specifically focusing on a point on that ellipse. An ellipse is generally represented as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

For a given point \((x_0, y_0)\) on the ellipse, the equation of the tangent line can be written using this relationship: \( \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1 \). This form helps us connect the dot between geometric representations and algebraic equations. In our example, the ellipse is \( \frac{x^2}{27} + y^2 = 1 \), and inserting the specific point \((3 \sqrt{3} \cos \theta, \sin \theta)\) into the tangent equation, you can get a specific tangent line described by the variables of \( \theta \).

This tangent line varies according to \( \theta \) which controls how the point moves along the curve of the ellipse.

Intercepts on Axes

The task of finding intercepts on the axes involves evaluating where the tangent line crosses the x-axis and y-axis. An intercept is a point where a line crosses the axis. So, for a tangent, the intercepts are crucial to understanding how the line lies in relation to the axes.

To find the x-intercept, you set \(y = 0\) in the tangent equation. In our context with the tangent to the ellipse, this gave us a result of \(x = 3 \sqrt{3} \sec \theta\). Similarly, setting \(x = 0\) helps find the y-intercept and gives \(y = \csc \theta\).

By calculating these intercepts, you get valuable coordinates showing how the line cuts through the Cartesian plane. This setup helped establish the sum of the intercepts needed for evaluating other properties such as minimizing expressions involving these intercepts.

Trigonometric Identities

Trigonometric identities are essential tools that make complex expressions manageable. They are equations involving trigonometric functions like sine, cosine, tangent, and their reciprocals. Here, the identities play a vital role in simplifying expressions with trigonometric terms.

For instance, in the derivative stage of minimizing our intercepts' expression, the identity \( \sec \theta = 1/\cos \theta \) and \( \csc \theta = 1/\sin \theta \) were used. These reciprocal identities help convert complex trigonometric terms into simpler expressions to manage, differentiate, and solve.

Moreover, identities like \(\sin^2 \theta + \cos^2 \theta = 1\) were inherently applied in forming those equations. Understanding and using these trigonometric identities offer elegance and ease in manipulating equations, which is crucial when looking to find critical points or specific values such as \(\theta\) for our exercise.

Minimum Value Problem

The problem of finding a minimum value for an expression derives from calculus, where you want to find the smallest value of a given function over a set of inputs. It's largely about optimization.

When deriving the sum of intercepts \(S(\theta) = 3 \sqrt{3} \sec \theta + \csc \theta\), the task was to determine the minimum value for different \(\theta\) values. To find this, we differentiate the function to identify where its rate of change equals zero — these are known as critical points.

Differentiation involves taking the derivative of \(S(\theta)\) and setting it to zero. This step determines where the function could potentially have its minimum value. Solving that, we found \(\theta = \frac{\pi}{6}\) as contributing to the minimum. Verification involved substituting this value back into the expression to ensure it does indeed present the most minimal sum of intercepts.

This involves calculus fundamentals applied through a series of logical approaches to find that ultimate least value of a function.