[Maths Class Notes] on Conic Sections Important Topics Pdf for Exam

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.