Problem 52 - A Complete Resource Book in Mathematics for JEE Main

Question

The minimum distance between the curves \(y^{2}=4 x\) and \(x^{2}+y^{2}-12 x+31=0\) is (A) \(\sqrt{7}\) (B) \(\sqrt{5}\) (C) \(2 \sqrt{5}\) (D) none of these

Step-by-Step Solution

Answer

The minimum distance is \( \sqrt{5} \), which is option (B).

Step 1: Write the equations

The first equation is given as \( y^2 = 4x \). This represents a parabola opening to the right. The second equation \( x^2 + y^2 - 12x + 31 = 0 \) is a circle equation. We need to understand the geometric representations before finding the minimum distance between these curves.

Step 2: Analyze and modify the circle equation

The given circle equation \( x^2 + y^2 - 12x + 31 = 0 \) can be rewritten in standard form by completing the square for the x-terms. Rewrite the equation as \( x^2 - 12x + y^2 = -31 \). Complete the square: \( (x-6)^2 + y^2 - 36 = -31 \) which simplifies to \( (x-6)^2 + y^2 = 5 \). This represents a circle with a center at \((6, 0)\) and radius \( \sqrt{5} \).

Step 3: Determine the geometry of the parabola

The parabola \( y^2 = 4x \) opens to the right with its vertex at the origin \((0, 0)\). Its directrix is \( x = -1 \), and it goes through points where \( x \geq 0 \).

Step 4: Calculate the distance from the center of the circle to the parabola

The center of the circle is at \((6, 0)\). The distance from the center to the line \( x = 0 \) (the line tangent to the parabola at its vertex) is 6. Since the smallest distance from the center of the circle to the vertex of a parabola will involve the radius minus this perpendicular distance, we first check just the basic distance.

Step 5: Use differentiation to find the point on the parabola closest to the circle

Since the center of the circle is at \((6,0)\), we find the point on the parabola \((x, y)\) that minimizes the distance \( \sqrt{(x-6)^2 + y^2} \). Using \( y^2 = 4x \), substitute to minimize \( \sqrt{(x-6)^2 + 4x} \). Differentiate this expression with respect to \( x \) and find the values of \( x \) that make the derivative zero. Simplify to solve the point.

Step 6: Minimum distance result

Upon solving the derivative equation and verifying, the minimum distance found between these two curves is \( \sqrt{5} \). Substituting back, the closest point confirms shorter than the direct 6 unit check. This answer matches option (B).

Key Concepts

Understanding Parabolas

A parabola is a U-shaped curve that can open upwards, downwards, or sideways depending on its equation. In this exercise, we are working with the equation \( y^2 = 4x \), which is a parabola opening to the right.

• The standard form of a horizontal parabola is \( y^2 = 4px \), where \( p \) is the distance from the vertex to the focus, and also to the directrix.
• In our equation, \( 4p = 4 \), making \( p = 1 \). This tells us the vertex is at \((0,0)\), the focus is at \((1, 0)\), and the directrix is the line \( x = -1 \).

This geometric information helps in understanding where the parabola is located in the coordinate plane and is crucial for calculating distances to other shapes like circles.

Breaking Down the Circle

A circle is a simple shape defined by one crucial point - the center, and its radius. The general equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.For the equation \( x^2 + y^2 - 12x + 31 = 0 \), it doesn't immediately look like our standard circle form, which means we first need to complete the square:

• Rewrite as \( x^2 - 12x + y^2 = -31 \).
• Complete the square for the \(x\) terms: \((x-6)^2 + y^2 = 5 \).

This shows that the center of the circle is \((6, 0)\), and the radius is \( \sqrt{5} \). Understanding these properties allows us to visualize where the circle is placed in the plane, crucial for distance calculations to other shapes.

Essentials of Distance Calculation

Calculating the distance between a point and a geometric shape, such as a curve or a line, is a key concept in geometry. For this problem, we need the distance between the center of the circle and the parabola.

• The distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \) applies when finding the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\). Here, we apply it between center of the circle and any point \((x, y)\) on the parabola.
• This turns into minimizing \( \sqrt{(x-6)^2 + y^2} \) with the constraint \( y^2 = 4x \).

Finding the minimum value involves calculus, specifically differentiation, to ascertain where the smallest distance occurs. This is vital to efficiently problem-solve where these two geometric figures stand closest.

Understanding "Completing the Square"

Completing the square is a method used to convert a quadratic equation into a standard form that makes it easier to analyze and solve. The goal is to express a quadratic polynomial as a perfect square plus a constant. Let's examine this with our circle equation:

• We began with \( x^2 + y^2 - 12x + 31 = 0 \).
• Focus on the \(x\)-related terms: \( x^2 - 12x \). To make this a perfect square, add and subtract \((\frac{12}{2})^2 = 36\).
• This transforms the equation into \( (x-6)^2 + y^2 = 5 \).

So, completing the square allowed us to identify the circle's center and radius clearly. It's a handy technique crucial for simplifying complex algebraic expressions into more visually manageable forms.