Problem 126 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A chord is drawn through the focus of the parabola \(y^{2}=6 x\) such that its distance from the vertex of this parabola is \(\frac{\sqrt{5}}{2}\), then its slope can be: (a) \(\frac{\sqrt{5}}{2}\) (b) \(\frac{\sqrt{3}}{2}\) (c) \(\frac{2}{\sqrt{5}}\) (d) \(\frac{2}{\sqrt{3}}\)

Step-by-Step Solution

Answer

The slope of the chord is \(\frac{\sqrt{5}}{2}\).

Step 1: Identify the vertex and focus of the parabola

The given parabola is \(y^2 = 6x\). For a parabola in the form \(y^2 = 4ax\), the vertex is at \((0,0)\) and the focus is at \((a,0)\). Here, \(4a = 6\), so \(a = \frac{3}{2}\). Thus, the vertex is \((0, 0)\) and the focus is \(\left(\frac{3}{2}, 0\right)\).

Step 2: Equation of the chord and distance from vertex

Let the chord be a line with an equation \(y = mx + c\). Since this chord passes through the focus \(\left(\frac{3}{2}, 0\right)\), substituting these values gives us \(0 = m\left(\frac{3}{2}\right) + c\), which leads to \(c = -\frac{3m}{2}\).

Step 3: Calculate distance from the vertex to the chord

The distance \(d\) of the line \(y = mx + c\) from the origin \((0,0)\) is given by \(d = \frac{|c|}{\sqrt{1 + m^2}}\). Substituting \(c = -\frac{3m}{2}\), we get:\[\frac{3m/2}{\sqrt{1 + m^2}} = \frac{\sqrt{5}}{2}\].

Step 4: Solve for the slope \(m\)

Square both sides:\[\left(\frac{3m}{2\sqrt{1 + m^2}}\right)^2 = \left(\frac{\sqrt{5}}{2}\right)^2\]\[\frac{9m^2}{4(1 + m^2)} = \frac{5}{4}\]Cross multiply:\[9m^2 = 5 + 5m^2\]Solving this, we get:\[4m^2 = 5\]\[m^2 = \frac{5}{4}\]Thus, \(m = \pm \frac{\sqrt{5}}{2}\).

Step 5: Compare solutions with given options

The possible values for the slope \(m\) are \(\frac{\sqrt{5}}{2}\) and \(-\frac{\sqrt{5}}{2}\). These correspond to one of the multiple-choice options (Option a).

Key Concepts

Focus of Parabola

The focus of a parabola is a crucial point that defines its shape and position. For the parabola given in the exercise, represented by the equation \( y^2 = 6x \), we can determine the focus using a standard form of the parabola equation, \( y^2 = 4ax \). Here, \( a \) represents the distance from the vertex to the focus along the x-axis.
To find the focus:

• Identify \( 4a = 6 \) from the equation.
• Solve for \( a \), which is \( a = \frac{6}{4} = \frac{3}{2} \).
• Thus, the focus is at the point \( \left( \frac{3}{2}, 0 \right) \).

The focus is where all the reflected lines from the parabola converge, influencing the parabola's orientation.

Distance from Vertex

The vertex of a parabola is often seen as the starting point for measuring anything related to the parabola's geometry. For the parabola \( y^2 = 6x \), the vertex is located at the origin \((0,0)\).
The distance concept comes into play when dealing with a line's position in relation to the parabola. In our case, a chord drawn through the focus creates a line whose distance from the vertex is a specific value, given as \( \frac{\sqrt{5}}{2} \).
To calculate this distance,

• Use the formula for the distance from a point to a line: \( d = \frac{|c|}{\sqrt{1 + m^2}} \).
• Here, \( c \) is derived from the line equation as \( -\frac{3m}{2} \).
• This equation ensures that the chord remains at the specified distance from the vertex.

Slope of a Line

The slope of a line is a measure of its steepness, defined as the ratio of vertical change to horizontal change between two points on the line. For a line \( y = mx + c \), \( m \) is the slope.
In the given exercise, the line or chord passes through the focus of the parabola. Hence, the value of \( m \), attributed to the slope, is pivotal in maintaining the chord's specific distance from the vertex.
To find this slope,

• Set up the equation \( \frac{3m/2}{\sqrt{1 + m^2}} = \frac{\sqrt{5}}{2} \).
• Solve the derived quadratic equation, \( m^2 = \frac{5}{4} \).
• The possible slopes become \( m = \pm \frac{\sqrt{5}}{2} \), indicating two potential directions for the line.

This dual possibility represents lines that can slope upwards or downwards.

Vertex of Parabola

The vertex of a parabola is the point where the parabola changes direction, often considered the 'tip' or 'turning point'. In standard form equations like \( y^2 = 4ax \), the vertex is located at \((0,0)\).
For our parabola, \( y^2 = 6x \), the vertex remains at the origin. This vertex helps in setting positions for other significant features such as focus and directrix.
Key roles of the vertex include:

• Serving as a reference point for the parabola's symmetry.
• Acting as the midpoint for focal chords like the one in the problem.
• Providing a basis for measuring distances such as the one given for the chord, \( \frac{\sqrt{5}}{2} \).

Understanding the vertex's position is critical for mastering quadratic planes and parabolic shapes.