Problem 66 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation of the directrix of the parabola \(y^{2}+4 y+\) \(4 x+2=0\) is: (A) \(x=-1\) (B) \(x=1\) (C) \(x=-\frac{3}{2}\) (D) \(x=\frac{3}{2}\)

Step-by-Step Solution

Answer

The equation of the directrix is \( x = \frac{3}{2} \) (Option D).

Step 1: Rewrite Equation in Standard Form

First, we need to rearrange the given equation to match the standard form of a parabola equation. The given equation is \( y^2 + 4y + 4x + 2 = 0 \). Let's rearrange it: \( y^2 + 4y = -4x - 2 \).

Step 2: Complete the Square for the y-terms

To simplify the equation, we complete the square for the \( y \) terms. Start with \( y^2 + 4y \). Add and subtract \(4\) (i.e., \( (\frac{4}{2})^2 = 4\)) inside the equation: \( y^2 + 4y + 4 - 4 = -4x - 2 \). This becomes \( (y + 2)^2 - 4 = -4x - 2 \).

Step 3: Simplify the Equation

Now convert the equation into standard form by moving constants: \( (y + 2)^2 = -4x + 2 \). Simplify the constants to \( (y + 2)^2 = -4(x - \frac{1}{2}) \).

Step 4: Identify Parabola Parameters

This equation \( (y+2)^2 = -4(x-\frac{1}{2}) \) represents a parabola that opens to the left. The vertex form \( (y-k)^2 = 4a(x-h) \) shows that the vertex \((h, k)\) is \((\frac{1}{2}, -2)\) and \( 4a = -4 \) implies \( a = -1 \).

Step 5: Find the Equation of the Directrix

The equation for the directrix of a parabola in the form \( (y-k)^2 = 4a(x-h) \) is \( x = h - a \). Since \( h = \frac{1}{2} \) and \( a = -1 \), the directrix is \( x = \frac{1}{2} - (-1) = \frac{1}{2} + 1 = \frac{3}{2} \).

Key Concepts

Equation of a Parabola

Parabolas are symmetrical curves and one of the most recognizable shapes in mathematics. They have a special equation that describes their shape and position on a coordinate plane. The standard form of a parabola centered around the x-axis is:

• Vertical parabolas: \( y = ax^2 + bx + c \)
• Horizontal parabolas: \( x = ay^2 + by + c \)

This form helps to describe how the parabola opens and its overall position. In these forms, the coefficients \(a\), \(b\), and \(c\) play crucial roles in determining the parabola's width, direction, and position of its vertex.
For example, in the problem given, we have a horizontal parabola because the x is on one side by itself after manipulation. Rewriting the initial equation: \[ y^2 + 4y + 4x + 2 = 0 \] aids in determining this as a horizontal parabola when compared to the general formula.

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square form. This technique is particularly useful for solving quadratic equations and rewriting equations of parabolas.
Here's a simple way to understand completing the square:

• First, identify the quadratic terms, such as \(y^2 + 4y\).
• Next, take half of the coefficient of the linear term \(b\), square it, and add and subtract this square within the equation to preserve equality.
• In our example: \((\frac{4}{2})^2 = 4\).

Thus, when you add and subtract 4, it becomes:\[ y^2 + 4y + 4 - 4 = 0 \]leading to:\[ (y + 2)^2 - 4 \]
Completing the square makes it simpler to transform the equation into vertex form and identify critical points of the parabola.

Parabola Vertex Form

The vertex form is a way of expressing the equation of a parabola to easily identify its vertex and direction. It looks like this for a parabola opening horizontally:\[ (y-k)^2 = 4a(x-h) \]Where:

• \((h, k)\) is the vertex of the parabola.
• \(a\) is a coefficient that determines the direction and width of the parabola.

In the given exercise, after completing the square and rearranging the equation, we transformed it into:\[ (y + 2)^2 = -4(x - \frac{1}{2}) \]From this, we can quickly ascertain:

• The vertex of the parabola is \((\frac{1}{2}, -2)\).
• The parabola opens to the left since \(a = -1\) is negative.

This form is not only efficient for identifying the vertex but is also crucial for finding other features such as the focus and directrix, which govern the geometric properties of the parabola.