Problem 155 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let the line \(y=m x\) and the ellipse \(2 x^{2}+y^{2}=1\) intersect at a point \(P\) in the first quadrant. If the normal to this ellipse at \(P\) meets the co- ordinate axes at \(\left(-\frac{1}{3 \sqrt{2}}, 0\right)\) and \((0, \beta)\), then \(\beta\) is equal to: (a) \(\frac{2 \sqrt{2}}{3}\) (b) \(\frac{2}{\sqrt{3}}\) (c) \(\frac{2}{3}\) (d) \(\frac{\sqrt{2}}{3}\)

Step-by-Step Solution

Answer

The value of \(\beta\) is \(\frac{2 \sqrt{2}}{3}\), so the answer is option (a).

Step 1: Understand the Intersection

The line \( y = mx \) intersects with the ellipse \( 2x^2 + y^2 = 1 \). By substituting \( y = mx \) into the ellipse equation, we obtain: \( 2x^2 + (mx)^2 = 1 \) or \( 2x^2 + m^2x^2 = 1 \). Simplify it to \((2 + m^2)x^2 = 1\). Thus \(x^2 = \frac{1}{2 + m^2}\), giving \(x = \frac{1}{\sqrt{2 + m^2}}\), and correspondingly, \( y = \frac{m}{\sqrt{2 + m^2}} \). The point of intersection \(P\) is \(\left(\frac{1}{\sqrt{2 + m^2}}, \frac{m}{\sqrt{2 + m^2}}\right)\).

Step 2: Find the Slope of the Normal

First, find the gradient of the tangent to the ellipse at \(P\). For the ellipse \( 2x^2 + y^2 = 1 \), implicitly differentiate with respect to \(x\) to get: \( 4x + 2yy' = 0 \), giving \( y' = -\frac{2x}{y} \). At point \(P\), \( y'= -\frac{2 \cdot \frac{1}{\sqrt{2 + m^2}}}{\frac{m}{\sqrt{2 + m^2}}} = -\frac{2}{m} \). The slope of the normal, which is the negative reciprocal of the tangent's slope, is \(\frac{m}{2}\).

Step 3: Use the Normal Slope to Find Intercepts

The normal line's equation is \( y - \frac{m}{\sqrt{2 + m^2}} = \frac{m}{2}\left(x - \frac{1}{\sqrt{2 + m^2}}\right) \). Set \(y = 0\) to find x-intercept: \(0 - \frac{m}{\sqrt{2 + m^2}} = \frac{m}{2}\left(x - \frac{1}{\sqrt{2 + m^2}}\right)\). Solving gives \( x = -\frac{1}{3\sqrt{2}} \). Now, set \(x = 0\) to find y-intercept at \(0 = \left(0 - \frac{1}{\sqrt{2 + m^2}}\right)\frac{m}{2} + \frac{m}{\sqrt{2 + m^2}}\), determining \( y = \beta\).

Step 4: Relate Intercepts to Solve for \(\beta\)

By equating the x-intercept found to \(-\frac{1}{3\sqrt{2}}\) and using the normal's slope, derive \(3\sqrt{2} = 3\sqrt{2}m(2 + m^2) = 2\), then \(m^2 = \frac{4}{3}\) and \(m = \frac{2}{\sqrt{3}}\). Substitute \(m\) back to obtain y-coordinates \(\beta: \frac{m}{\sqrt{2 + m^2}} = \beta \). Simplify to get \( \frac{2/\sqrt{3}}{\sqrt{2 + 4/3}} \equiv \beta = \frac{2\sqrt{2}}{3}\).

Step 5: Select the Correct Option

Compare your found value for \(\beta\) with the multiple-choice options: (a) \(\frac{2 \sqrt{2}}{3}\), confirming this to be exact. Thus, the answer to the problem is option (a).

Key Concepts

Intersection of line and curve

When dealing with the intersection of a line and a curve, like an ellipse, it's crucial to determine the exact points where the two meet. In this case, we have the line given by the equation \( y = mx \) and the ellipse represented by \( 2x^2 + y^2 = 1 \). By substituting the expression for \( y \) from the line equation into the ellipse equation, you simplify the problem to one involving a single variable rather than two.

To find the intersection points, perform the following steps:

• Substitute \( y = mx \) into \( 2x^2 + y^2 = 1 \), transforming it into \( 2x^2 + m^2x^2 = 1 \).
• Combine terms to get \( (2 + m^2)x^2 = 1 \).
• Solve for \( x \) to find \( x = \frac{1}{\sqrt{2 + m^2}} \).
• Use \( y = mx \) to find \( y = \frac{m}{\sqrt{2 + m^2}} \).
• The intersection point \( P \) is at \( \left( \frac{1}{\sqrt{2 + m^2}}, \frac{m}{\sqrt{2 + m^2}} \right) \).

Understanding these steps to find the intersection helps you clearly locate the point \( P \) on both the line and the ellipse.

Normal to a curve

The normal to a curve at a given point is a line that is perpendicular to the tangent at that point. In this exercise, finding the normal line involves using the tangent's slope information at point \( P \) on the ellipse.

Here's how the process works:

• The equation of the ellipse is \( 2x^2 + y^2 = 1 \).
• Use implicit differentiation to find the derivative \( y' \), which represents the slope of the tangent line.
• Differentiating gives \( 4x + 2yy' = 0 \), so \( y' = -\frac{2x}{y} \).
• Substitute the coordinates of point \( P \) to get the slope of the tangent at \( P \), which is \( -\frac{2}{m} \).
• The slope of the normal is the negative reciprocal of the tangent's slope; therefore, it is \( \frac{m}{2} \).

Once you have the slope of the normal, you can find its equation and further analyze its intercepts with the coordinate axes.

Implicit differentiation

Implicit differentiation is an essential technique used to find the derivative of a function when it is not given in the standard format \( y = f(x) \). In this exercise, implicit differentiation helps us find the slope of the tangent to the ellipse.

The steps to perform implicit differentiation for the ellipse \( 2x^2 + y^2 = 1 \) are as follows:

• Differentiate each term individually with respect to \( x \): for \( 2x^2 \), use the power rule to get \( 4x \).
• For \( y^2 \), apply the chain rule: differentiate \( y^2 \) as \( 2y \cdot y' \), where \( y' \) is the derivative of \( y \) with respect to \( x \).
• Set the derivative of the equation equal to zero: \( 4x + 2yy' = 0 \).
• Solve for \( y' \) to get \( y' = -\frac{2x}{y} \).

This process allows you to determine the rate of change or slope of the tangent line at any point on the curve, which is critical for finding the equation of the normal line.

Quadrants in coordinate geometry

Coordinate geometry divides the Cartesian plane into four quadrants. The positioning of points in these quadrants is important for understanding their geometric properties. This exercise specifies that the intersection occurs in the first quadrant, which is essential for the values of \( x \) and \( y \).

Here's what each quadrant signifies about a point:

• First Quadrant: Both \( x \) and \( y \) are positive. Here, point \( P \) in this quadrant means \( \frac{1}{\sqrt{2 + m^2}} > 0 \) and \( \frac{m}{\sqrt{2 + m^2}} > 0 \).
• Second Quadrant: \( x \) is negative and \( y \) is positive.
• Third Quadrant: Both \( x \) and \( y \) are negative.
• Fourth Quadrant: \( x \) is positive and \( y \) is negative.

Knowing how to identify and work with these quadrants helps simplify the analysis of curves and their intersections with other geometric figures.