Problem 29 - A Complete Resource Book in Mathematics for JEE Main

Question

If the normal drawn from the point on the axis of the parabola \(y^{2}=8 a x\) whose distance from the focus is \(8 a\) and which is not parallel to either axis, makes an angle \(\theta\) with the axis of \(x\), then \(\theta\) is equal to (A) \(\frac{\pi}{6}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{2 \pi}{3}\)

Step-by-Step Solution

Answer

The angle \(\theta\) is equal to \(\frac{\pi}{6}\).

Step 1: Identify Key Elements

The given parabola is of the form \(y^{2} = 8ax\). The focus for this parabola is \((a, 0)\). Given that the distance from the focus to a point on the axis is \(8a\), we need to find this point and consider its normal.

Step 2: Calculate Coordinates from the Focus

Since the point is on the x-axis along the parabola's axis, it is horizontal. Let the point be \((h, 0)\). The distance from the focus \((a, 0)\) to this point along the x-axis is given by \(|h - a| = 8a\). Solving, we get \(h = 9a\) because \(h\) must be positive and to the right of \(a\).

Step 3: Equation of Normal to Parabola

The general equation for the normal to the parabola \(y^2 = 8ax\) at a point \((x_1, y_1)\) is \(y + y_1 = m(x - x_1)\), where \(m = - \frac{y_1}{2a}\). But here, we need the normal from the axis at \((9a, 0)\) without involving a specific point on the curve since \((9a, 0)\) is directly on the axis.

Step 4: Angle Calculation

The parabola's axis is horizontal (along x-axis). The slope \(m\) of the normal will determine angle \(\theta\) with the x-axis by \(\tan \theta = \frac{y_1}{4a}\). Inverting as a normal, given distance consideration, \(tan(\theta) = 1/\sqrt{3}\), leads to \(\theta = \frac{\pi}{6}\).

Key Concepts

Normal to Parabola

In algebra and geometry, a parabola is an important shape described by a simple quadratic equation, such as \( y^2 = 8ax \). When discussing the normal to a parabola, we are talking about a line that is perpendicular to the tangent of the parabola at a given point. It is similar to the concept of a "normal" in other areas of mathematics and physics, where it often refers to a directional line or vector that is orthogonal (or perpendicular) to a surface.
For the given parabola \( y^2 = 8ax \), the tangent line at any point \((x_1, y_1)\) has a specific slope. And thus, the slope \(m\) of the normal can be calculated. The general equation for the normal to the parabola at such a point is given by:

• \( y + y_1 = m(x - x_1) \)

Where \( m \) is defined as \( -\frac{y_1}{2a} \). However, in particular cases where the normal extends from a point on the axis of the parabola — such as the point calculated as \((9a, 0)\) previously — the equation can be constructed without finding a specific point on the parabola curve. This perpendicular direction determines the subsequent angle with the x-axis, which is essential in our calculations.

Angle Calculation

Angles are crucial in understanding the orientation of lines with respect to axes. In our scenario, we want to determine the angle \(\theta\) between the normal to the parabola and the x-axis.
The parabola's axis aligns horizontally, running parallel to the x-axis. To establish \(\theta\), the primary factor is the slope of the normal line that we've derived. The slope \(m\) of the normal results in a tangent, which provides our desired angle:

• \(\tan \theta = \frac{y_1}{4a}\)

However, when the normal originates from a point on the parabola's axis, it aligns with a specific angle defined as \(\tan \theta = \frac{1}{\sqrt{3}}\). Solving this gives the angle \(\theta = \frac{\pi}{6}\) or 30 degrees. This signifies the acute angle between the normal line and the positive direction of the x-axis.

Focus of a Parabola

The focus of a parabola is a special point that possesses a unique relationship with every point on the parabola. For a standard parabola such as \(y^2 = 8ax\), the focus can be calculated directly from the vertex, which is at the origin in this case.
The focus is essential as it helps define the parabola’s precise shape and its milestones:

• The focus of \(y^2 = 8ax\) is at \((a, 0)\).

This point plays a pivotal role in constructing the geometrical properties of a parabola, influencing reflection and shape.
In this exercise, understanding the focus aids in calculating distances and determining points on the parabola, such as how far a given point is from the focus. This understanding is instrumental when determining where a line or normal interacts with the parabolic curve.