Class 11 RS Aggarwal Chapter-23 Ellipse Solutions - Free PDF Download
FAQs on RS Aggarwal Class 11 Solutions Chapter - 23 Ellipse for Fast Learning
1. How do I find the key properties like foci, vertices, and eccentricity from the standard equation of an ellipse as per RS Aggarwal Chapter 23?
To find the key properties of an ellipse from its standard equation (e.g., x²/a² + y²/b² = 1), follow these steps:
- Identify the Major Axis: Compare the denominators. If the denominator of x² (which is a²) is larger than that of y² (b²), the major axis is horizontal. If b² is under x² and a² is under y², it's vertical.
- Find 'a' and 'b': Take the square root of the denominators to find the values of 'a' (semi-major axis) and 'b' (semi-minor axis).
- Calculate 'c': Use the formula c² = a² - b² to find the distance 'c' from the centre to a focus.
- Determine Properties: Use these values to find the vertices (±a, 0 or 0, ±a), foci (±c, 0 or 0, ±c), and eccentricity (e = c/a).
2. What are the main types of problems covered in the RS Aggarwal Class 11 Maths solutions for Chapter 23, Ellipse?
The RS Aggarwal solutions for Class 11 Chapter 23 primarily guide students through solving the following types of problems:
- Finding the equation of an ellipse when given specific conditions, such as its foci, vertices, or the length of the latus rectum.
- Calculating the key parameters (foci, vertices, major/minor axes, eccentricity) from a given standard equation of an ellipse.
- Solving problems involving the latus rectum of an ellipse.
- Determining the equation of a tangent and normal to an ellipse at a given point.
- Finding the position of a point with respect to an ellipse.
3. How can RS Aggarwal Solutions for Class 11 Ellipse help in solving complex problems step-by-step?
The RS Aggarwal solutions for the Ellipse chapter are designed to build problem-solving skills. They provide a structured, step-by-step methodology for each question. This helps you understand how to apply the correct formulas, break down complex problems into manageable parts, and verify your steps. Following these solutions helps reinforce the correct approach to ensure you can tackle similar problems accurately in exams.
4. How do you derive the equation of an ellipse if only its foci and the length of the major axis are provided?
This is a common problem type in Chapter 23. You can solve it using the following method:
- Find the Centre: The centre of the ellipse (h, k) is the midpoint of the line segment connecting the two foci.
- Find 'a': The length of the major axis is given as 2a. Halve this value to get 'a'.
- Find 'c': The distance from the centre to either focus is 'c'. Calculate this distance using the coordinates.
- Find 'b²': Use the fundamental relationship c² = a² - b². Rearrange it to b² = a² - c² to calculate the value of b².
- Write the Equation: Substitute the values of a² and b² into the standard form of the ellipse equation, orienting the major axis correctly based on the position of the foci.
5. Why is the concept of the latus rectum important when solving problems on the ellipse in RS Aggarwal?
The latus rectum is a crucial parameter because it defines the width of the ellipse at its focus. Its length, given by the formula 2b²/a, is a unique property of an ellipse. In problem-solving, it often acts as a key piece of information that connects the semi-major axis 'a' and semi-minor axis 'b'. You can use the length of the latus rectum to find one of the axes if the other is known, which is essential for determining the final equation of the ellipse.
6. What is the key difference in the method for solving problems for a horizontal ellipse versus a vertical ellipse?
The key difference lies in the orientation of the major axis, which changes the coordinates of the foci and vertices. For a horizontal ellipse (x²/a² + y²/b² = 1), the major axis is along the x-axis, so the foci are at (±c, 0) and vertices are at (±a, 0). For a vertical ellipse (x²/b² + y²/a² = 1), the major axis is along the y-axis, placing the foci at (0, ±c) and vertices at (0, ±a). While the calculation of a, b, and c using c² = a² - b² remains the same, correctly identifying the orientation is the first and most critical step to ensure the coordinates are correct.
7. How do you find the equation of a tangent to an ellipse at a specific point (x₁, y₁) using the methods in Chapter 23?
The solutions in RS Aggarwal demonstrate a direct method. For a standard ellipse with the equation x²/a² + y²/b² = 1, the equation of the tangent at a point (x₁, y₁) on the ellipse is found by making the following replacements in the ellipse's equation:
- Replace x² with xx₁
- Replace y² with yy₁
This gives the equation of the tangent as xx₁/a² + yy₁/b² = 1. This formula provides a quick and accurate way to solve for the tangent without needing complex calculus-based derivations.











