Problem 184 - A Complete Resource Book in Mathematics for JEE Main

Question

The normal to the curve, \(x^{2}+2 x y-3 y^{2}=0\), at \((1,1):\) \([2015 \mid\) (A) meets the curve again in the second quadrant. (B) meets the curve again in the third quadrant. (C) meets the curve again in the fourth quadrant. (D) does not meet the curve again.

Step-by-Step Solution

Answer

The normal meets the curve again in the fourth quadrant. Option (C).

Step 1: Differentiate the Curve

We start by differentiating the given curve implicitly with respect to x. The curve equation is:\[ x^2 + 2xy - 3y^2 = 0 \] Differentiating with respect to x, we have:\[ 2x + 2y + 2x \frac{dy}{dx} - 6y \frac{dy}{dx} = 0 \]Simplifying, it gives:\[ (2x - 6y) \frac{dy}{dx} = -2x - 2y \]Thus, the slope of the tangent at any point \((x, y)\) is:\[ \frac{dy}{dx} = \frac{-2x - 2y}{2x - 6y} \]

Step 2: Calculate the Tangent Slope at (1,1)

Using the point \((1,1)\), substitute into the slope formula to find the slope of the tangent:\[ \frac{dy}{dx} = \frac{-2(1) - 2(1)}{2(1) - 6(1)} = \frac{-4}{-4} = 1 \]So, the slope of the tangent at \((1, 1)\) is 1.

Step 3: Find the Normal Slope and Equation

The normal is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent slope.Since the tangent slope is 1, the normal slope is -1.The equation of the normal at \((1, 1)\) is:\[ y - 1 = -1(x - 1) \]Simplifying, we get:\[ y = -x + 2 \]

Step 4: Find the Intersection of Normal with the Curve

Substitute the normal equation \(y = -x + 2\) back into the original curve to find the intersection points:\[ x^2 + 2x(-x + 2) - 3(-x + 2)^2 = 0 \]Which simplifies to:\[ x^2 - 2x^2 + 4x - 3(x^2 - 4x + 4) = 0 \]Simplify further:\[ x^2 - 2x^2 + 4x - 3x^2 + 12x - 12 = 0 \]Combine terms:\[ -4x^2 + 16x - 12 = 0 \]Divide by -4:\[ x^2 - 4x + 3 = 0 \]Factor:\[ (x - 3)(x - 1) = 0 \]So, \(x = 3\) or \(x = 1\).

Step 5: Confirm Meeting Point

We know that \((1,1)\) is our starting point, we substitute \(x = 3\) back to find its y-coordinate:\[ y = -3 + 2 = -1 \]Thus, the normal meets the curve again at \((3, -1)\), which is in the fourth quadrant.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when you have relationships between two variables that are not explicitly expressed as functions of each other. In our exercise, we have the equation of a curve given as \[ x^2 + 2xy - 3y^2 = 0 \] which involves both \( x \) and \( y \) intermingling.

Unlike explicit differentiation where \( y \) is isolated, here, we differentiate both sides of the equation with respect to \( x \). This requires us to also apply the chain rule when dealing with terms involving \( y \), as \( y \) itself depends on \( x \). So, when we differentiate \( xy \), we treat \( y \) as a function \( y(x) \), and use the product rule:

• Differentiating \( xy \) gives: \( rac{d}{dx} (xy) = x rac{dy}{dx} + y \).
• For terms like \( y^2 \), the chain rule applies: \( rac{d}{dx} (y^2) = 2y \frac{dy}{dx} \).

These rules are essential when differentiating such relationships, leading to an expression for \( \frac{dy}{dx} \) which represents the slope of the tangent line at any point on the curve.

Quadrants

In the coordinate plane, the four quadrants are key to understanding where points lie. These quadrants are defined by the positive and negative directions of the x and y axes:

• First Quadrant: Both x and y are positive.
• Second Quadrant: x is negative, y is positive.
• Third Quadrant: Both x and y are negative.
• Fourth Quadrant: x is positive, y is negative.

In our exercise, we determine where the normal line intersects the curve again. We first found our normal line equation, \( y = -x + 2 \), and solved for intersection points. One known intersection is at \((1, 1)\) in the first quadrant.

However, when solving the equations, we identified another meeting point at \((3, -1)\), which indeed lies in the fourth quadrant. The x-coordinate is positive and the y-coordinate is negative, thus confirming its location in this quadrant.

Intersection Points

Finding intersection points between two graphs involves solving them simultaneously. This means finding common solutions (\(x,y\) pairs) that satisfy both equations.

For example, to find where the normal line \( y = -x + 2 \) intersects our curve again, we need to substitute \( y = -x + 2 \) back into the original curve equation:\[ x^2 + 2x(-x + 2) - 3(-x + 2)^2 = 0 \]
By simplifying, we find a quadratic equation in \( x \):\[ x^2 - 4x + 3 = 0 \]This needs to be factored:\[ (x - 3)(x - 1) = 0 \]
The solutions \( x = 1 \) and \( x = 3 \) are found, meaning the curve and the normal line intersect at these x-values. As \((1,1)\) is our starting point, we check the other x-value \( x = 3 \), substitute back into the line equation to find \( y \), which results in the point \((3, -1)\). This satisfies both the normal and the original curve equation, confirming it as an intersection point.