Problem 96 - A Complete Resource Book in Mathematics for JEE Main

Question

If two tangents drawn from a point \(P\) to the parabola \(y_{1}^{2}\) \(=4 x\) are at right angles, then the locus of the point \(P\) is (A) \(2 x+1=0\) (B) \(x=-1\) (C) \(2 x-1=0\) (D) \(x=1\)

Step-by-Step Solution

Answer

Option (C) \(2x-1=0\) is the correct answer.

Step 1: Express the Equation of Tangents

For a parabola \( y^2 = 4ax \), the equation of the tangent from a point \( (x_1, y_1) \) outside the parabola can be written as \( yy_1 = 2a(x + x_1) \). Here, we have \( a = 1 \), because the given parabola is \( y^2 = 4x \). So, the tangent equation simplifies to \( yy_1 = 2(x + x_1) \).

Step 2: Condition for Perpendicular Tangents

The tangents from point \( P(x_1, y_1) \) to the parabola are perpendicular. For two lines to be perpendicular, the product of their slopes is \(-1\). For the tangents, the slopes derived from normal forms are \( m_1 = \frac{2}{y_1} \) and \( m_2 = \frac{-2}{y_1} \). Their product is \( m_1 \times m_2 = \frac{2}{y_1} \times \frac{-2}{y_1} = \frac{-4}{y_1^2} = -1 \).

Step 3: Finding Condition on \( y_1 \)

From \( \frac{-4}{y_1^2} = -1 \), it simplifies to \( y_1^2 = 4 \). Therefore, \( y_1 = \pm 2 \). This provides the necessary condition for perpendicular tangents from point \( P \).

Step 4: Determine the Locus of \( P \)

Since we need the locus of point \( P(x_1, y_1) \) for \( y_1 = \pm 2 \), we substitute into the equation of the parabola: \( y_1^2 = 4 \). So the equation simplifies as \( 2x = 1 \), which is \( x = \frac{1}{2} \).

Step 5: Match the Locus with Given Options

According to the options provided, \( 2x - 1 = 0 \) simplifies to \( x = \frac{1}{2} \), which corresponds to option (C).

Key Concepts

Equation of Tangent

To understand how tangents work with a parabola, let's start with the equation of a parabola: \( y^2 = 4ax \). When dealing with tangents, we're trying to find a line that just "touches" the parabola at a single point. If you have a point \((x_1, y_1)\) outside the parabola, the equation of the tangent line from that point is \( yy_1 = 2a(x + x_1) \). Here, the constant \(a\) is specific to the parabola's formula. In the case of the parabola \(y^2 = 4x\), we identify that \(a = 1\). Therefore, our tangent equation becomes \( yy_1 = 2(x + x_1) \).Understanding the tangent equation is crucial because it tells you exactly how a line from a given point can just touch the curve, and not cross it. This tangency condition is integral when calculating things like the slope of the tangent or in finding out more about the geometry of shapes like parabolas.

Locus

The term "locus" generally means a set of points that satisfy specific geometric conditions. When asked to find the locus, such as for a point \(P\) from which perpendicular tangents can be drawn to a parabola, you're essentially looking for all possible positions for that point under the given conditions.In the exercise, the parabola was \(y^2 = 4x\), and we are finding the locus of the point \(P\) where tangents from \(P\) are perpendicular. The mathematical journey led us to the point where \( y_1^2 = 4 \), meaning \( y_1 = \pm 2 \).By substituting \( y_1 = \pm 2 \) back into the parabola's equation, we determined \( 2x - 1 = 0 \) or equivalently \( x = \frac{1}{2} \) as the equation describing that locus. The locus equation collects all points \(P\) where these conditions are satisfied, essentially telling us where to "find" point \(P\) for the geometric scenario to hold true.

Perpendicular Lines

Perpendicular lines meet at a right angle, which is 90 degrees. In this context, two tangents are perpendicular if their slopes multiply to \(-1\). For any two lines in a coordinate plane, this perpendicularity condition can be mathematically represented as:

• If the slope of one line is \(m_1\) and the slope of the other is \(m_2\), then \(m_1 \times m_2 = -1\).

In our exercise concerning the parabola \(y^2 = 4x\), the slopes of the tangents are calculated to be \(\frac{2}{y_1}\) and \(\frac{-2}{y_1}\). So, \(m_1 \times m_2 = \frac{2}{y_1} \times \frac{-2}{y_1} = \frac{-4}{y_1^2}\).To ensure these tangents meet at a right angle, the equation \(\frac{-4}{y_1^2} = -1\) must be true. Solving this gives us \(y_1^2 = 4\) or \(y_1 = \pm 2\). These values ensure the product of the slopes remains constant at \(-1\), making the lines perpendicular. It’s this principle which helps to derive conclusions about the points from which the tangents are drawn.