Problem 174 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A stair-case of length \(l\) rests against a vertical wall and a floor of a room. Let P be a point on the stair-case, nearer to its end on the wall, that divides its length in the ratio \(1: 2\). If the stair-case begins to slide on the floor, then the locus of P is: (a) an ellipse of eccentricity \(\frac{1}{2}\) (b) an ellipse of eccentricity \(\frac{\sqrt{3}}{2}\) (c) a circle of radius \(\frac{1}{2}\) (d) a circle of radius \(\frac{\sqrt{3}}{2} l\)

Step-by-Step Solution

Answer

(b) an ellipse of eccentricity \( \frac{\sqrt{3}}{2} \).

Step 1: Define the position of point P on the staircase

The length of the staircase is divided by point P in the ratio 1:2. This indicates that one segment from the bottom end of the staircase (on the floor) to P is \(\frac{1}{3}\) of the entire length \(l\), and the other segment from P to the top end on the wall is \(\frac{2}{3}\) of the length \(l\).

Step 2: Use the Pythagorean theorem for movement

Let \(x\) be the distance from the wall to the base of the ladder on the floor, and \(y\) be the height from the floor to the top end of the ladder on the wall. Then, according to the Pythagorean theorem, we have: \(x^2 + y^2 = l^2\).

Step 3: Determine the coordinates of point P

Since P is \(\frac{1}{3}\) of the way up the ladder, its coordinates are \((\frac{x}{3}, \frac{2y}{3})\).

Step 4: Formulate the equation for the locus of P

Substitute \(x = 3X\) and \(y = \frac{3Y}{2}\) into the Pythagorean relation \(x^2 + y^2 = l^2\), so that \( (3X)^2 + \left( \frac{3Y}{2} \right)^2 = l^2\). This simplifies to: \[9X^2 + \frac{9Y^2}{4} = l^2\].

Step 5: Simplify the locus equation

To simplify the equation, divide by 9, yielding: \[X^2 + \frac{Y^2}{4} = \frac{l^2}{9}\]. This describes an ellipse centered at the origin with \(a^2 = \frac{l^2}{9}\) and \(b^2 = \frac{l^2}{36}\).

Step 6: Calculate the eccentricity of the ellipse

The eccentricity \(e\) of the ellipse is given by \(e = \sqrt{1 - \frac{b^2}{a^2}}\). Substituting \(b^2 = \frac{l^2}{36}\) and \(a^2 = \frac{l^2}{9}\), we get: \[e = \sqrt{1 - \frac{\frac{l^2}{36}}{\frac{l^2}{9}}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2} \].

Key Concepts

Ellipse

An ellipse is a type of curve on a plane surrounding two focal points. The sum of the distances to the two focal points is constant for every point on the ellipse. It's like a stretched-out circle. If you picture an elongated silhouette of an oval shape, that's roughly what an ellipse looks like.

Ellipses are significant in various areas of mathematics and physics. For example, they describe the orbits of planets around the sun. The general equation of an ellipse in its standard form is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. These two distances determine how wide and how tall the ellipse is.

The shape of the ellipse changes based on \(a\) and \(b\). If both are equal, we have a circle, which is actually a special case of an ellipse.

Eccentricity

Eccentricity is a number that determines how "stretched" an ellipse is. It is denoted by the symbol \(e\). For an ellipse, the eccentricity is defined as \( e = \sqrt{1 - \frac{b^2}{a^2}} \), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. This value ranges from 0 to 1.

- When \(e = 0\), we have a perfect circle.- As \(e\) approaches 1, the ellipse becomes more stretched and elongated.- For eccentricity values approaching 1, the ellipse appears almost like a line.

In our example, calculating the eccentricity of the ellipse where the locus of point \(P\) lies, we use the provided formula. Substituting the given values, we find that \( e = \frac{\sqrt{3}}{2} \), indicating an ellipse that is moderately stretched. Understanding eccentricity helps in visualizing and distinguishing between different types of ellipses.

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry, connecting the lengths of sides in a right-angled triangle. The theorem states: "In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides."

Mathematically, it appears as \( a^2 + b^2 = c^2 \), where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides. This theorem is enormously helpful for calculating distances and in many other problem-solving scenarios.

In the context of our exercise, we used the Pythagorean theorem to connect the distances from the ladder’s endpoints on the wall and floor. With \(x\) being the horizontal distance from the wall and \(y\) the vertical height on the wall, the relationship \(x^2 + y^2 = l^2\) stems directly from this important theorem. It's a versatile tool that facilitates our understanding of the relationship between different lengths in geometric figures.