Problem 100 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let a line \(y=m x(m>0)\) intersect the parabola, \(y^{2}=x\) at a point \(P\), other than the origin. Let the tangent to it at \(P\) meet the \(x\)-axis at the point \(Q\), If area \((\Delta O P Q)=4 \mathrm{sq}\). units, then \(m\) is equal to ___ .

Step-by-Step Solution

Answer

\(m = \frac{1}{2^{1/3}}\).

Step 1: Find Intersection Points

The line \(y = mx\) intersects the parabola \(y^2 = x\) at points \(P\) and \(O(0,0)\). Substituting \(y = mx\) into \(y^2 = x\) gives \((mx)^2 = x\). This simplifies to \(m^2x^2 = x\). For this to be true, either \(x = 0\) (the origin) or \(m^2x = 1\). Thus, \(x = \frac{1}{m^2}\) when \(x eq 0\). Substitute \(x = \frac{1}{m^2}\) into \(y = mx\) to find \(y = \frac{m}{m^2} = \frac{1}{m}\). Therefore, the coordinates of point \(P\) are \(\left(\frac{1}{m^2}, \frac{1}{m}\right)\).

Step 2: Calculate Slope of the Tangent at P

The slope of the parabola at a point \((x,y)\) can be found using implicit differentiation of \(y^2 = x\). Differentiating both sides gives \(2y \frac{dy}{dx} = 1\). Thus, \(\frac{dy}{dx} = \frac{1}{2y}\). At point \(P\), \(y = \frac{1}{m}\), so \(\frac{dy}{dx} = \frac{1}{2 \times \frac{1}{m}} = \frac{m}{2}\).

Step 3: Equation of the Tangent Line at P

The equation of the tangent line at \(P\) is given by the point-slope form: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is \(\left(\frac{1}{m^2}, \frac{1}{m}\right)\), and the slope \(m\) is \(\frac{m}{2}\). Therefore, the tangent line equation is \(y - \frac{1}{m} = \frac{m}{2}\left(x - \frac{1}{m^2}\right)\). Simplifying, we find \(y = \frac{m}{2}x - \frac{1}{2m} + \frac{1}{m}\).

Step 4: Find x-intercept of Tangent (Point Q)

Set \(y = 0\) in the tangent equation to find where it meets the \(x\)-axis: \(0 = \frac{m}{2}x - \frac{1}{2m} + \frac{1}{m}\). Solve for \(x\), which gives \(x = \frac{2}{m^2}\). Thus, point \(Q\) is \(\left(\frac{2}{m^2}, 0\right)\).

Step 5: Calculate Area of Triangle OPQ

The triangle \(\Delta OPQ\) has vertices \(O(0,0)\), \(P\left(\frac{1}{m^2}, \frac{1}{m}\right)\), and \(Q\left(\frac{2}{m^2}, 0\right)\). The area \(A\) of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), and \((x_3,y_3)\) is given by \(A = \frac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|\). For \(\Delta OPQ\), \(A = \frac{1}{2}\left|0\left(\frac{1}{m} - 0\right) + \frac{1}{m^2}(0 - 0) + \frac{2}{m^2}\left(\frac{1}{m} - 0\right)\right| = \frac{1}{2}\left|\frac{2}{m^3}\right| = \frac{1}{m^3}\).

Step 6: Solve for m

The problem states that the area \(A = 4\) square units. Therefore, \(\frac{1}{m^3} = 4\). Solve for \(m\) by taking the reciprocal and cube root: \(m^3 = \frac{1}{4}\). Thus, \(m = \frac{1}{\sqrt[3]{4}}\). Simplifying, \(m = \frac{1}{2^{1/3}}\).

Key Concepts

Intersection of Parabola and Line

When exploring the intersection of a parabola and a line, we typically aim to find the points where these two graphs meet on a coordinate plane. In this scenario, our line is defined by the equation \( y = mx \), where \( m > 0 \) indicates a line with a positive slope. The parabolic curve is described by \( y^2 = x \).
To determine the intersection points, we substitute the line equation into the parabola's equation. This substitution yields the equation \((mx)^2 = x\). Solving it, we find possibilities: either \( x = 0 \), which corresponds to the origin, or after simplifying, \( x = \frac{1}{m^2} \). The corresponding \( y \)-coordinate is determined by substituting back into the line equation resulting in \( y = \frac{1}{m} \). Hence, point \( P \) of intersection, aside from the origin, is \( (\frac{1}{m^2}, \frac{1}{m}) \).

• The parabola authentically represents conical sections or curves.
• Lines interjected with such curves require solving simultaneous equations.
• Every rise in \( m \) tends to pivot the line, altering intersection solutions.

Tangent to Parabola

A tangent line is unique as it just "touches" a curve at a particular point, offering insights into geometry's limits and rates of change. The parabola tangent at point \( P(\frac{1}{m^2}, \frac{1}{m}) \) implies a line with exactly one point of contact.
At any point on a parabola like \( y^2 = x \), the gradient or slope of the tangent can be determined by implicit differentiation. Differentiating yields \( 2y \frac{dy}{dx} = 1 \), simplifying to \( \frac{dy}{dx} = \frac{1}{2y} \). At our specific point \( P \), we find \( \frac{dy}{dx} = \frac{m}{2} \) since \( y = \frac{1}{m} \).

With the point-slope form of lines, \( y - y_1 = m(x - x_1) \), the equation of our tangent line becomes \( y - \frac{1}{m} = \frac{m}{2}(x - \frac{1}{m^2}) \). Simplified further, this presents \( y = \frac{m}{2}x - \frac{1}{2m} + \frac{1}{m} \).

• Derivatives give tangents' equations by providing slopes.
• Tangents only touch the curve without fully intersecting.
• Tangent lines reflect changing slopes based on curvature.

Area of Triangle in Coordinate Geometry

In coordinate geometry, calculating the area of a triangle helps bridge spatial concepts with algebra. Specifically, given points \( O(0,0) \), \( P(\frac{1}{m^2}, \frac{1}{m}) \), and \( Q(\frac{2}{m^2}, 0) \), we utilize the area formula for triangles formed by vertices on the plane.
The formula \( A = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \) provides a structured approach to finding area using coordinate points. For \( \triangle OPQ \), calculating the determinant simplifies to \( \frac{1}{m^3} \) after substituting in coordinates.

Given that the area equals 4 square units, \( \frac{1}{m^3} = 4 \). Solving for \( m \) involves taking the reciprocal: \( m^3 = \frac{1}{4} \), leading to \( m = \frac{1}{2^{1/3}} \).

• Coordinate geometry links algebraic expressions to geometric insights.
• The formula evaluates signed areas to rectify orientation issues.
• Analyzing triangle areas enhances understanding of object intersections in space.