Problem 95 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: A normal is drawn at a point \(P(x, y)\) of a curve. It meets the \(x\)-axis and the \(y\)-axis in points \(A\) and \(B\), respectively, such that \(\frac{1}{O A}+\frac{1}{O B}=1\), where \(O\) is the origin. The equation of such a curve passing through \((5,4)\) is \((x-1)^{2}+(y-1)^{2}=25\). Reason: \(O A=x+y \frac{d y}{d x}\) and \(O B=\frac{\left(x+y \frac{d y}{d x}\right)}{\frac{d y}{d x}}\)

Step-by-Step Solution

Answer

The assertion is true, and the reason is also correctly calculated, hence valid.

Step 1: Understand the problem

We have a curve given by the equation \((x-1)^2 + (y-1)^2 = 25\). A normal to the curve at any given point \((x, y)\) intersects the x-axis and y-axis at points A and B, respectively. We want \(\frac{1}{OA} + \frac{1}{OB} = 1\), where O is the origin. We also need to show that the assertion and reason are true given the derived conditions.

Step 2: Equation of the Curve

Recognize that the given equation \((x-1)^2+(y-1)^2 = 25\) is the equation of a circle centered at \((1, 1)\) with radius 5.

Step 3: Find the equation of the normal

The slope of the tangent to the curve is \(\frac{dy}{dx} = -\frac{x-1}{y-1}\). Therefore, the slope of the normal is \(\frac{y-1}{x-1}\). The equation of the normal to the curve at \((x,y)\) is then \(\frac{y-1}{x-1} = \frac{y-y_1}{x-x_1}\), which simplifies to \( (y-1)(x-x_1) = (x-1)(y-y_1)\).

Step 4: Find points A and B

To find point A where the normal intersects the x-axis, set \(y=0\) in the normal equation. To find point B where it intersects the y-axis, set \(x=0\) in the same equation.

Step 5: OA and OB Expressions

After solving the equations for A \((x_0, 0)\) and B \((0, y_0)\), compute \(OA = |x_0|\) and \(OB = |y_0|\). These are the points measured from the origin.

Step 6: Validate the Assertion

Plug the values of OA and OB into \(\frac{1}{OA} + \frac{1}{OB} = 1\). If the equation holds true, then the assertion is verified.

Step 7: Analyze the Reason

Use the derived expressions to check if \( OA = x + y \frac{dy}{dx} \) and \( OB = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx}} \) hold true as described in the reason.

Step 8: Final verification

If the reasons match the calculated values and the assertion is conclusively shown to be true, then both are validated.

Key Concepts

Circle Equation

In geometry, a circle's equation helps us understand its position and size on a coordinate plane. The general form, \((x-h)^2 + (y-k)^2 = r^2\), represents a circle centered at \((h, k)\) with radius \(r\). In our exercise, the circle is given by \((x-1)^2 + (y-1)^2 = 25\).
This equation tells us that the circle is centered at \((1, 1)\) and has a radius of 5.
To visualize, imagine drawing a perfect circle around the point \((1, 1)\) with every point on the circle being exactly 5 units away from the center.

• The "h" and "k" in the equation are the x and y coordinates of the center.
• The squared terms ensure each point of the circle satisfies the constant distance from the center.

Understanding the circle equation is crucial as it affects other concepts like tangents and normals, especially when finding intersection points on coordinate planes.

Normal to a Curve

A normal to the curve at a point is a perpendicular line to the tangent at that point. For any point \((x, y)\) on the curve \((x-1)^2 + (y-1)^2 = 25\), the tangent and normal lines have specific slopes:

• The slope of the tangent is derived as \( \frac{dy}{dx} = -\frac{x-1}{y-1} \), meaning it's the rate at which y changes relative to x.
• The slope of the normal, being perpendicular, is the negative reciprocal, \( \frac{y-1}{x-1} \).

This normal intersects the axes to form a line of interest in many geometric and analytic problems. Understanding these slopes helps when constructing and solving geometric properties, including perpendicularity and linear intersections.

Coordinate Geometry

Coordinate geometry combines algebra and geometry to solve geometric issues using coordinate points. In this section, it simplifies complex curves and lines through algebraic expressions.
This practice involves finding key properties like the intersection of normals with axes by setting up equations and solving for variables such as the points A and B mentioned in the problem.
Coordinate geometry allows students to translate geometric figures like circles and lines into a form where they can use algebraic techniques, making it easier to solve intricate problems.

• It systematically applies formulas to calculate better the distances, slopes, and measures along the x and y axes, like \(OA\) and \(OB\).
• It aids in visualizing and analyzing the positions of points, lines, and curves graphically.

Effectively using coordinate geometry streamlines approaching and resolving geometric problems through algebraic methods.

Intersection Points

Finding intersection points involves identifying where curves and lines meet in a coordinate plane. In this problem, you identify where the normal intersects the x-axis (A) and the y-axis (B). This requires substituting specific values into the normal equation:

• Letting \(y = 0\) finds where it crosses the x-axis.
• Letting \(x = 0\) locates the y-axis intersection.

This concept is crucial for solving geometric problems as these intersection points often represent significant, actionable points tied to the circle equation. When finding \(OA\) and \(OB\), these segments from the origin to the intersection points demonstrate crucial understanding intersections influence on curve analysis.
The relationship \(\frac{1}{OA} + \frac{1}{OB} = 1\) in this exercise illustrates how different axes intersections contribute to a particular problem scenario, creating specific geometric arrangements.