Problem 117 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Tangent and normal are drawn at \(\mathrm{P}(16,16)\) on the parabola \(\mathrm{y}^{2}=16 \mathrm{x}\), which intersect the axis of the parabola at A and \(\mathrm{B}\), respectively. If \(\mathrm{C}\) is the centre of the circle through the points \(\mathrm{P}, \mathrm{A}\) and \(\mathrm{B}\) and \(\angle \mathrm{CPB}=\theta\), then a value of \(\tan \theta\) is: (a) 2 (b) 3 (c) \(\frac{4}{3}\) (d) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of \(\tan \theta\) is 2.

Step 1: Find the Equation of the Tangent

The given parabola is \(y^2 = 16x\). The point \(P(16, 16)\) lies on this parabola. The general equation of the tangent to the parabola at point \((x_1, y_1)\) is \(yy_1 = 2a(x + x_1)\), where \(a = 4\) for this parabola. Substituting \((x_1, y_1) = (16, 16)\), the equation becomes \(16y = 8(x + 16)\), or \(2y = x + 16\). Rearranging gives \(x = 2y - 16\).

Step 2: Find the Point on the X-axis where the Tangent Intersects

To find the intersection of the tangent with the x-axis, set \(y = 0\) in the tangent equation \(x = 2y - 16\). Solving gives \(x = -16\). Therefore, tangent intersects the axis at point \(A(-16, 0)\).

Step 3: Find the Equation of the Normal

The slope of the tangent is \(\frac{1}{2}\), so the slope of the normal is the negative reciprocal, \(-2\). The equation of the normal at \((16, 16)\) is \(y - 16 = -2(x - 16)\). Simplifying gives \(y = -2x + 48\).

Step 4: Find the Intersect Point of the Normal with the X-axis

To find the intersection of the normal with the x-axis, set \(y = 0\) in the normal equation \(y = -2x + 48\). Solving gives \(x = 24\). Therefore, the normal intersects the axis at point \(B(24, 0)\).

Step 5: Find the Center of the Circle Passing through A, B, and P

The center \(C\) of the circle through points \(A(-16, 0)\), \(B(24, 0)\), and \(P(16, 16)\) is the midpoint of the segment joining \(A\) and \(B\). This midpoint is computed as \(\left(\frac{-16 + 24}{2}, 0\right) = (4, 0)\).

Step 6: Determine \(\angle CPB\) and Calculate \(\tan \theta\)

We know that \(C(4, 0)\), \(P(16, 16)\), and \(B(24, 0)\). Calculate the slope of \(CP\) as \(\frac{16 - 0}{16 - 4} = \frac{4}{3}\) and the slope of \(PB\) as \(\frac{16 - 0}{16 - 24} = -2\). \(\tan \theta\) is given by the formula for the angle between two lines, \(\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\), where \(m_1 = \frac{4}{3}\) and \(m_2 = -2\). Substituting these into the formula gives \(\tan \theta = \frac{\left| \frac{4}{3} + 2 \right|}{\left| 1 - \left(\frac{4}{3}\right)(-2) \right|}\), which simplifies to \(2\).

Step 7: Select the Correct Answer

Based on the calculations, \(\tan \theta = 2\), which matches option (a).

Key Concepts

Equation of Tangent

When you have a curve like a parabola, a tangent line is a straight line that touches the curve at exactly one point. At this point, the tangent has the same direction as the curve. For the parabola given by the equation \( y^2 = 16x \), and at point \( P(16, 16) \), we use the formula for the tangent. This formula is \( yy_1 = 2a(x + x_1) \) where \( (x_1, y_1) \) is the point of contact. Here, the parameter \( a = 4 \). By substituting the coordinates of point \( P \) into the formula, we find:

• The equation becomes \( 16y = 8(x + 16) \)
• After rearranging, it simplifies to \( x = 2y - 16 \)

This is the equation of the tangent line at point \( P \). To find where this tangent intersects the x-axis, set \( y = 0 \). Solving gives \( x = -16 \), so it crosses the x-axis at \( A(-16, 0) \). This point of intersection is key in further solving the problem.

Equation of Normal

The normal line to a curve at a given point is perpendicular to the tangent line at that point. For our parabola, with the tangent slope found to be \( \frac{1}{2} \), the normal line has a slope that's the negative reciprocal, which is \( -2 \). The general formula for the line through a point \((x_1, y_1)\) with slope \( m \) is \( y - y_1 = m(x - x_1) \). Using point \( P(16, 16) \) and the normal slope:

• The equation becomes \( y - 16 = -2(x - 16) \)
• This simplifies to \( y = -2x + 48 \)

This is the equation of the normal line through point \( P \). Setting \( y = 0 \) to find where it intersects the x-axis gives \( x = 24 \), leading to point \( B(24, 0) \). This intersection is part of constructing a circle and calculating angles related to the problem.

Intersection with Axes

The intersections of lines with the coordinate axes are crucial in analyzing geometric relations and solving for unknowns. In this exercise, the parabola's tangent and normal lines intersect the x-axis at different points, which are essential for forming a circle.

• For the tangent, setting \( y = 0 \) in \( x = 2y - 16 \) gave point \( A(-16, 0) \).
• For the normal, setting \( y = 0 \) in \( y = -2x + 48 \) yielded point \( B(24, 0) \).

These points are used to find the center of the circle passing through points \( A, B, \text{and } P(16,16) \). This center \( C \) is determined as the midpoint of the line segment \( AB \) which is \( (4,0) \). Understanding these intersections is key to solving for \( \tan \theta \) in the problem, as it involves these geometric relationships.