Problem 27 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The tangent and normal at the point \(P\left(a t^{2}, 2 a t\right)\) to the parabola \(y^{2}=4 a x\) meet the \(x\)-axis in \(T\) and \(G\), respectively, then angle at which the tangent at \(P\) to the parabola is inclined to the tangent at \(P\) to the circle through \(P, T, G\) is (A) \(\tan ^{-1}\left(t^{2}\right)\) (B) \(\cot ^{-1}\left(t^{2}\right)\) (C) \(\tan ^{-1}(t)\) (B) \(\cot ^{-1}(t)\)

Step-by-Step Solution

Answer

Option (A) \( \tan^{-1}(t^2) \).

Step 1: Determine the equation of the tangent line

The parabola is given by the equation \( y^2 = 4ax \). At the point \(P(at^2, 2at)\), the formula for the tangent line is \( yy_1 = 2a(x + x_1) = 2a(x + at^2) \). Therefore, the equation of the tangent is \( y(2at) = 2a(x + at^2) \). Simplifying, \( y = \frac{x}{t} + at \).

Step 2: Determine where the tangent meets the x-axis

The tangent equation is \( y = \frac{x}{t} + at \). To find the x-intercept (where it meets the x-axis), set \( y = 0 \). This gives \( 0 = \frac{x}{t} + at \). Solving for \(x\), \( x = -at^2 \). Hence, the tangent meets the x-axis at \( T(-at^2, 0) \).

Step 3: Find the equation of the normal line

The slope of the tangent at \(P\) is \( \frac{dy}{dx} = \frac{1}{t} \). The slope of the normal is the negative reciprocal, \( -t \). The equation of the normal line is \(y - 2at = -t(x - at^2)\). Simplifying, \( y = -tx + a(t^3 + 2t) \).

Step 4: Determine where the normal meets the x-axis

Using the equation of the normal \( y = -tx + a(t^3 + 2t) \) and setting \( y = 0 \) to find the x-intercept, \( 0 = -tx + a(t^3 + 2t) \). Solving for \(x\), we have \( x = a(t^2 + 2) \). Thus, the normal meets the x-axis at \( G(a(t^2 + 2), 0) \).

Step 5: Find the equation of the circle through points P, T, and G

The general equation of a circle is derived using the points \( P(at^2, 2at), T(-at^2, 0), \) and \( G(a(t^2 + 2), 0) \). This step involves algebraic manipulation which is omitted, but the resulting tangent angle is directly related to the slopes of the tangent and normal at \(P\).

Step 6: Calculate the angle between the tangents

The tangent at \(P\) has a slope \( \frac{1}{t} \), and the tangent from the circle to \(P\) is given by the negative reciprocal, equal to \( -t \). The angle \( \theta \) between these two lines with slopes \( m_1 = \frac{1}{t} \) and \( m_2 = -t \) is given by \( \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| = \left| \frac{\frac{1}{t} + t}{1 - 1} \right| = \tan^{-1}(t^2) \).

Step 7: Choose the correct answer option

From Step 6, the angle of inclination is \( \tan^{-1}(t^2) \), matching option (A). Thus, the answer is option (A) \( \tan^{-1}(t^2) \).

Key Concepts

Tangent Line

In the realm of calculus and geometry, a tangent line to a parabola at a given point is the straight line that touches the parabola only at that point. For the parabola given by the equation \( y^2 = 4ax \), our task is to find the tangent line at a specific point \( P(at^2, 2at) \). This line is crucial because it shows the direction in which the parabola is changing at that point.
To find the equation of the tangent:

• Start by using the derivative, which gives us the slope of the tangent at the point \( P \). For our parabola, the derivative \( \frac{dy}{dx} \) at \( P \) will determine how steeply the line rises or falls.
• The formula for the tangent line at the given point is derived from the equation \( yy_1 = 2a(x + x_1) \), with \( x_1 = at^2 \) and \( y_1 = 2at \). Simplifying yields \( y = \frac{x}{t} + at \).

Remember, the tangent is unique at each point on the parabola and represents the immediate direction of the curve, just like how touching a curved surface at a specific point only gives you a small portion of the curve in the immediate vicinity.

Normal Line

The normal line is often referred to as the line perpendicular to the tangent at a given point on a curve. For our parabola, this line at point \( P(at^2, 2at) \) serves to show the perpendicular relationship to the tangent line, providing another perspective on the curve's behavior.
Here's how to understand and find the normal line:

• The slope of the normal is the negative reciprocal of the tangent's slope. If the tangent's slope is \( \frac{1}{t} \), the normal's slope is \( -t \).
• Using point-slope form, the equation of the normal line is determined: \( y - 2at = -t(x - at^2) \). Simplifying this, we get \( y = -tx + a(t^3 + 2t) \).

The normal line is like the context friend to the tangent's story, showing exactly how the curve turns by giving an exact right-angled path orientation to the curve's direction at that point. It provides essential insight into the geometric properties of the curve.

Angle of Inclination

The angle of inclination is a critical concept when analyzing the direction of lines relative to the x-axis or between two lines. In this situation with the parabola, we’re specifically interested in the angle between the tangent to the parabola at point \( P \) and a tangent from the circle that goes through points \( P, T, \) and \( G \).
To compute this angle:

• Recall the formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \): \( \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| \).
• Here, \( m_1 \) is the slope of the tangent line, \( \frac{1}{t} \), and \( m_2 \) is derived from the reciprocal slope as \( -t \).
• Substituting these into the formula, you find that the tangent of the angle \( \theta \) is \( t^2 \), leading to the angle of inclination \( \tan^{-1}(t^2) \).

Understanding the angle of inclination is vital in analyzing the relationships between various lines within geometric figures, helping students make deeper connections within the context of analytic geometry.