Must-Know Concepts in Conic Sections for Exams
Conic Sections topics covered in this article will give you a vivid description of the Class 11 Chapter Conic chapter. These solutions will widely help during the exams and also for its preparation. These solutions contain essential question answers and solvable questions that may come in the examination. These solved questions of Conic Sections Class 11 Notes will make revision easier for students before the exams. Conic Sections Class 11 Notes is an easy and scoring chapter. During the exams, if you follow these solved questions, it will clear all doubts. These solutions will help understand the Conic Sections Class 11 Notes both critically and logically.
Short Description of Conic Section
Conic Sections Class 11 Notes are explained in simple steps here. These explanations are easy to understand and clear all doubts at once. Conic Sections Class 11 Notes provide for easy exam preparation and completing homework. The solution provided will help to solve any question related to this chapter. Conic Sections Class 11 Notes will make the subject interesting and fun to learn. Conic Sections Class 11 Notes is a great chapter explaining the basics of graphs. Conic Sections Class 11 Notes is an easy and scoring chapter. Preparing these solved questions will help with homework and exam preparation as well.
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Importance of Conic Sections
Conic Sections Class 11 Notes helps to prepare for exams without any worries. It will make students work much more manageably.
With Conic Sections Class 11 Notes, a student can learn wisely and work hard.
Make a routine and allot time for different subjects. Routine makes learning fun and values time.
A key to good scores is to revise the previous year’s question paper from Conic Sections Class 11 Notes.
Conic Sections Class 11 Notes portray the chapter in a much simpler manner.
It provides all the short answer type questions and Solved examples for the student guide.
Solved Examples
1) Find the centre and the radius of the circle x2 + y2 + 8x + 10y – 8 = 0
Solution: The given equation is as follows
or, x2 + y2 + 8x + 10y – 8 = 0
or, (x2 + 8x) + (y2 + 10y) = 8
By completing the squares within the parenthesis
or, (x2 + 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25
or, (x + 4)2 + (y + 5)2 = 49
or, [x – (– 4)]2 + [y – (–5)]2 = 72
Comparing with the standard form: a = -4, b = -5 and t = 7
Hence, the given circle has centre at (– 4, –5) and the radius is 7.
2) A circle passes through the points (2, – 2), and (3, 4). It has the centre on the line x + y = 2. Find the equation of the circle.
Solution: Let the equation of the circle be (x – a)2 + (y – b)2 = t2.
the circle passes through the points (2, –2) and (3, 4)is given.
Hence,
(2 – a)2 + (–2 – b)2 = t2 (1)
and (3 – a)2 + (4 – b)2 = t2 (2)
Also, given that the centre lies on the line x + y = 2.
or, a + b = 2 (3)
Solving the equations (1), (2), (3), we get
h = 0.7, k = 1.3 and t2 = 12.58
Hence, the equation of the given circle is
(x – 0.7)2+ (y – 1.3)2 = 12.58.
3) Find the equation of the circle with center (–3, 2) and radius 4.
Solution: Given,
Centre = (a, b) = (-3, 2)
Radius = r = 4
Therefore, the equation of the required circle is
⇒ (x – a)2 + (y – b)2 = t2
4) Find the centre and the radius of the circle with the equation x2 + y2 + 8x + 10y – 8 = 0
Solution: The given equation is as follows
or, x2 + y2 + 8x + 10y – 8 = 0
or, (x2 + 8x) + (y2 + 10y) = 8
By completing it
⇒ (x2 + 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25
i.e. (x + 4)2 + (y + 5)2 = 49
i.e. [x – (– 4)]2 + [y – (–5)]2 = 72
Comparing with the standard form, a = -4, b= -5 and t = 7, here t is the radius.
Hence, the given circle has a centre at (– 4, –5) and a radius of 7.
FAQs on Essential Conic Sections Topics and Their Importance
1. Why is the chapter on Conic Sections considered important for the Class 11 Maths exam for the 2025-26 session?
Conic Sections is a crucial chapter in Coordinate Geometry for the Class 11 final exams. It holds significant weightage and is fundamental for understanding concepts in both mathematics and physics. Questions from this chapter frequently test a student's ability to apply standard formulas to find parameters like focus, vertex, and eccentricity, making it a high-scoring topic if the core concepts and their applications are clear.
2. What are the standard equations of the four conic sections that are essential for the CBSE Class 11 exam?
For the Class 11 Maths exam, you must be thorough with the standard forms of the four conic sections. Mastering these is an important first step. The equations are:
- Circle: The standard equation is (x - h)² + (y - k)² = r², where (h, k) is the centre and r is the radius. When the centre is the origin, the equation simplifies to x² + y² = r².
- Parabola: The four standard forms are y² = 4ax, y² = -4ax, x² = 4ay, and x² = -4ay.
- Ellipse: The standard equations are x²/a² + y²/b² = 1 (major axis along x-axis) and x²/b² + y²/a² = 1 (major axis along y-axis), for a > b.
- Hyperbola: The standard equations are x²/a² - y²/b² = 1 (transverse axis along x-axis) and y²/a² - x²/b² = 1 (transverse axis along y-axis).
3. How do you find the focus, directrix, and latus rectum of a standard parabola like y² = 16x?
This is a frequently asked type of question. To find the properties of the parabola y² = 16x, follow these steps:
- Step 1: Compare with Standard Form. The given equation matches the standard form y² = 4ax. By comparing, we get 4a = 16, which means a = 4.
- Step 2: Find the Focus. The focus of this type of parabola is at (a, 0). Therefore, the focus is (4, 0).
- Step 3: Find the Directrix. The equation of the directrix is x = -a. So, the directrix is x = -4.
- Step 4: Calculate the Length of the Latus Rectum. The length of the latus rectum is given by the formula 4a. In this case, it is 16.
4. What is the significance of eccentricity (e), and how does it help differentiate between the conic sections in an exam?
Eccentricity (e) is a crucial parameter that defines the shape of any conic section. It represents the ratio of the distance of a point on the curve from the focus to its perpendicular distance from the directrix. Understanding its value is a higher-order thinking skill (HOTS) because it uniquely identifies the conic, which is vital for solving complex problems:
- For a Circle, e = 0.
- For a Parabola, e = 1.
- For an Ellipse, 0 < e < 1.
- For a Hyperbola, e > 1.
5. From an exam perspective, what are the most common types of important questions from the Conic Sections chapter?
For the CBSE Class 11 exams, important questions from Conic Sections typically fall into these categories:
- 1-Mark Questions: These usually test basic knowledge, like finding the centre/radius of a circle from its equation or identifying the type of conic.
- 3-Mark Questions: These require you to find key properties like the focus, vertex, directrix, and axes of a given parabola, ellipse, or hyperbola.
- 5-Mark Questions: These are often application-based, requiring you to derive the equation of a conic when conditions like foci, vertices, or eccentricity are given.
6. How can a student avoid common mistakes when identifying horizontal and vertical ellipses or hyperbolas during an exam?
A common trap is confusing the orientation of an ellipse or hyperbola. Here’s a simple way to avoid errors:
- For an Ellipse (x²/a² + y²/b² = 1): The major axis is determined by the larger denominator (which is always a²). If a² is under x², the ellipse is horizontal. If a² is under y², it is vertical.
- For a Hyperbola (x²/a² - y²/b² = 1 or y²/a² - x²/b² = 1): The transverse axis is determined by the positive term. If the x² term is positive, the hyperbola is horizontal. If the y² term is positive, it is vertical. The denominator of the positive term is always a².
7. What are the key differences in the properties and equations of an ellipse and a hyperbola that are important for solving problems?
Understanding the core differences between an ellipse and a hyperbola is essential for applying the correct formulas. The main distinctions are:
- Defining Condition: For an ellipse, the sum of the distances from any point on it to its two foci is constant. For a hyperbola, the difference of these distances is constant.
- Standard Equation: An ellipse's equation uses addition (e.g., x²/a² + y²/b² = 1), while a hyperbola's equation uses subtraction (e.g., x²/a² - y²/b² = 1).
- Key Relationship: For an ellipse, the relationship between the semi-major axis (a), semi-minor axis (b), and distance from centre to focus (c) is c² = a² - b². For a hyperbola, it is c² = a² + b².
