Problem 99 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The point on the curve \(3 x^{2}-4 y^{2}=72\) which is nearest to the line \(3 x+2 y+1=0\) is (A) \((6,-3)\) (B) \((6,3)\) (C) \((-6,3)\) (D) \((-6,-3)\)

Step-by-Step Solution

Answer

The point (6, -3) on the hyperbola is closest to the line.

Step 1: Understand the Problem

We need to find the point on the hyperbola defined by the equation \(3x^2 - 4y^2 = 72\) that is closest to the line \(3x + 2y + 1 = 0\). This involves finding the minimum distance between a point on the hyperbola and the given line.

Step 2: Express Distance Formula

The distance \(d\) between a point \((x_1, y_1)\) and a line \(Ax + By + C = 0\) is given by \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\). Here, \(A = 3\), \(B = 2\), and \(C = 1\), so the distance from \((x_1,y_1)\) on the hyperbola to the line is \(\frac{|3x_1 + 2y_1 + 1|}{\sqrt{13}}\).

Step 3: Use Lagrange Multipliers

To minimize the distance, use Lagrange multipliers. The objective function is the distance, and the constraint is the hyperbola equation \(3x^2 - 4y^2 - 72 = 0\). Set the Lagrangian \(\mathcal{L}(x, y, \lambda) = 3x + 2y + 1 + \lambda(3x^2 - 4y^2 - 72)\) and find the critical points by taking its partial derivatives with respect to \(x, y, \lambda\), and set them to zero.

Step 4: Solve Partial Derivatives

Calculate the partial derivatives: \(\frac{\partial \mathcal{L}}{\partial x} = 3 + \lambda(6x) = 0\), \(\frac{\partial \mathcal{L}}{\partial y} = 2 - \lambda(8y) = 0\), and \(\frac{\partial \mathcal{L}}{\partial \lambda} = 3x^2 - 4y^2 - 72 = 0\). Solve these equations together with the constraint to find possible \((x,y)\) pairs.

Step 5: Resolve the Equations

From \(3 + 6\lambda x = 0\), we have \(\lambda = -\frac{1}{2x}\). From \(2 - 8\lambda y = 0\), we have \(\lambda = \frac{1}{4y}\). Equating the two expressions for \(\lambda\) gives \(-\frac{1}{2x} = \frac{1}{4y}\). Solving, we get \(-2y = x\). Substitute \(x = -2y\) into the constraint equation to solve for \(x\) and \(y\).

Step 6: Solve for x and y

Substitute \(x = -2y\) into the hyperbola equation: \(3(-2y)^2 - 4y^2 = 72\), which simplifies to \(12y^2 - 4y^2 = 72\), or \(8y^2 = 72\). Hence, \(y^2 = 9\), so \(y = \pm 3\). Using \(x = -2y\), when \(y = 3\), \(x = -6\); when \(y = -3\), \(x = 6\). We have points \((-6, 3)\) and \((6, -3)\).

Step 7: Compare Distances

Calculate and compare distances from each point to the line \(3x + 2y + 1 = 0\) using the distance formula. For \((-6, 3)\), the distance is \(\frac{|3(-6) + 2(3) + 1|}{\sqrt{13}}\); for \((6, -3)\), the distance is \(\frac{|3(6) + 2(-3) + 1|}{\sqrt{13}}\). Calculate these to find the smaller distance.

Key Concepts

Hyperbola

A hyperbola is a fascinating type of conic section that you encounter in analytic geometry. You can think of it as a curve on a plane, where each point on the hyperbola is such that the absolute difference of its distances to two fixed points, known as the foci, remains constant. The equation of a hyperbola is similar in form to that of an ellipse, but with a key difference. Its equation typically looks like this:

• \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)

This particular structure creates two distinct open curves, or branches, which extend infinitely. In the given problem, the hyperbola is represented by the equation \(3x^2 - 4y^2 = 72\). You can transform this into the standard form by dividing each term by 72, which shows that it has branches opening symmetrically along either the x-axis or y-axis.
This symmetry is a critical feature, and understanding it can help you solve many problems involving hyperbolas effectively.
In some problems, you need to find the points on the hyperbola closest to a given line, often requiring optimization techniques, like Lagrange multipliers, to find such points.

Distance Formula

The distance formula is a fundamental tool in geometry that helps determine the distance between a point and a line. This formula is key in solving problems related to finding the shortest distance, especially in coordinate geometry, such as in our exercise. The formula for calculating the distance \(d\) from a point \((x_1, y_1)\) to a line given by \(Ax + By + C = 0 \) is expressed as:

• \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)

Understanding this formula is vital as it provides the shortest distance possible from the point to the line. It's the orthogonal, or perpendicular, distance.
Using it, you can compare distances from various points to a line and solve for minimum or maximum distances. In this exercise, after finding potential points on the hyperbola, we calculate the exact distances to the given line to determine which point is nearer. This comparison helps in making a selection among multiple possibilities, ensuring that the solution is both precise and based on accurate calculations.

Lagrange Multipliers

Lagrange multipliers are a powerful optimization tool used in calculus to find the maxima and minima of functions subject to constraints. While this may sound complex, it's just a strategic method to ensure that the solution satisfies specific conditions while optimizing a particular objective function.
The problem with finding the point on the hyperbola closest to a line involves minimizing the distance function subject to the hyperbola's equation. To achieve this, you define a Lagrangian function, combining the distance formula as the objective and the hyperbola equation as a constraint. The Lagrangian function is:

• \( \mathcal{L}(x, y, \lambda) = 3x + 2y + 1 + \lambda(3x^2 - 4y^2 - 72) \)

Next, find solutions by calculating partial derivatives with respect to each variable \((x, y, \lambda)\) and equating them to zero.
This procedure helps in identifying critical points where the function could have an extreme value.
This analytical method yields efficient solutions, particularly helpful when dealing with complex algebraic expressions associated with hyperbolas and lines.