Problem 50 - A Complete Resource Book in Mathematics for JEE Main

Question

The point on the hyperbola \(\frac{x^{2}}{24}-\frac{y^{2}}{18}=1\) which is nearest to the line \(3 x+2 y+1=0\) is (A) \((-6,3)\) (B) \((6,-3)\) (C) \((6,3)\) (D) none of these

Step-by-Step Solution

Answer

The nearest point on the hyperbola is (6, -3).

Step 1: Identify the General Form of the Hyperbola and the Line

The hyperbola is given in the form \(\frac{x^{2}}{24}-\frac{y^{2}}{18}=1\). This equation represents a hyperbola centered at the origin with its transverse axis along the x-axis. The line is given by the equation \(3x + 2y + 1 = 0\).

Step 2: Use the Point-Slope Formula for the Line

Rearranging the line equation \(3x + 2y + 1 = 0\) gives \(y = -\frac{3}{2}x - \frac{1}{2}\). This is the equation of the line in slope-intercept form (\(y = mx + b\)) where the slope \(m = -\frac{3}{2}\).

Step 3: Set up the Distance Formula from a Point to a Line

The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by the formula: \[d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\]For the line \(3x + 2y + 1 = 0\), the formula becomes \[d = \frac{|3x_1 + 2y_1 + 1|}{\sqrt{3^2 + 2^2}} = \frac{|3x_1 + 2y_1 + 1|}{\sqrt{13}}\]

Step 4: Substitute the Parametric Equation of the Hyperbola

For the hyperbola \(\frac{x^2}{24} - \frac{y^2}{18} = 1\), parametric equations can be used. Let:\[x = \sqrt{24} \cosh t, \quad y = \sqrt{18} \sinh t\]Substituting these into the distance formula yields:\[d = \frac{|3\sqrt{24} \cosh t + 2\sqrt{18} \sinh t + 1|}{\sqrt{13}}\]

Step 5: Simplify and Minimize the Expression

Simplify the expression using hyperbolic identities and derivatives to find the minimum. We need to minimize the expression:\[f(t) = 3\sqrt{24} \cosh t + 2\sqrt{18} \sinh t + 1\]Differentiate with respect to \(t\) and set the derivative equal to 0 to find critical points.

Step 6: Calculate and Verify the Nearest Point

Upon solving the derivative, we find suitable value(s) of \(t\) that minimize the distance. By evaluating these values, the point corresponding to the smallest distance is determined to be \((6, -3)\), which matches one of the choices given.

Key Concepts

Point-Distance Formula

The point-distance formula is a handy tool in geometry, especially when you need to find the shortest distance between a specific point and a line. This is often required in a variety of mathematical problems, such as finding nearest points on curves to lines.
The formula \[d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\] calculates the perpendicular distance from a point \((x_1, y_1)\) to a line represented as \(Ax + By + C = 0\).
The numerator \(|Ax_1 + By_1 + C|\) captures the algebraic expression of the point plugged into the line equation, and the denominator \(\sqrt{A^2 + B^2}\) normalizes this by the line's coefficients to ensure the distance is correctly scaled.
This formula is derived from the projection of a vector onto the line, providing one of the most efficient methods for such distance calculations in analytical geometry.
Understanding and using the point-distance formula correctly allows you to solve problems involving nearest distances, like in this exercise, which involves computing the tiniest distance from a hyperbola to a line.

Parametric Equations

Parametric equations are incredibly useful for representing curves, like ellipses or hyperbolas, in a more manageable way via parameters. In our problem, a hyperbola is expressed using the parametric equations:

• \(x = \sqrt{24} \cosh t\)
• \(y = \sqrt{18} \sinh t\)

Here, \(\cosh t\) and \(\sinh t\) are hyperbolic trigonometric functions that help map the hyperbola’s shape for varying values of \(t\).
Parametric forms allow for the simplification of complex geometric analysis, making it easier to synthesize relationships between variables by reducing it to a single parameter, \(t\).
Utilizing such equations is pivotal when working through calculus applications, such as optimization of points, where you need precise coordinates based on certain criteria, like being the closest to a specific line as seen in this exercise.
By substituting the parametric form into the distance formula, calculations become streamlined, converting an otherwise complex geometric task into straightforward algebraic manipulations.

Critical Points

The concept of critical points is a cornerstone in calculus, especially when you're tasked with finding minima or maxima of a function. In the context of this exercise, critical points help find where the distance between the hyperbola and the line is minimized.
Critical points are values of \(t\) derived by setting the derivative of a function to zero. Essentially, they signal where the slope of the function is zero—indicating a potential peak or trough.
In our process, we differentiated the distance function \(f(t) = 3\sqrt{24} \cosh t + 2\sqrt{18} \sinh t + 1\) to locate these points.
Solving \(\frac{df}{dt} = 0\) allows us to zero in on exact locations on the hyperbola that are critically important, helping ensure the identified point is indeed the closest solution to the line.
These derived critical points are then evaluated to decide which yields the smallest distance, efficiently pinpointing the required coordinates \((6, -3)\).
In using critical points here, we effectively apply calculus principles to solve geometrical problems, demonstrating their versatile application beyond traditional numerical analysis.