NCERT Solutions for Class 11 – Complete Chapter-wise Study Material
11. Conic Sections is one of the most important chapters in the Class 11 Mathematics English NCERT Solutions curriculum. This chapter plays a significant role in helping students build a strong conceptual foundation while preparing for school examinations, class tests, unit tests, half-yearly examinations, annual examinations, and CBSE board assessments. The chapter has been carefully designed according to the latest NCERT syllabus, making it an essential part of every student's study plan.
The 11. Conic Sections - Class 11 Mathematics English NCERT Solutions available on ATP Education explain every question in a simple, accurate, and step-by-step manner. Each answer is prepared according to the latest CBSE guidelines so that students can understand the concepts clearly without confusion. Whether you are completing your homework, revising before examinations, or strengthening your understanding of the subject, these solutions provide reliable academic support throughout your learning journey.
One of the biggest advantages of studying 11. Conic Sections is that it helps students understand important concepts, definitions, examples, and textbook exercises in an organized way. Instead of memorizing answers, students learn how to develop logical thinking, improve analytical skills, and write well-structured answers in examinations. This chapter also helps improve problem-solving ability and encourages conceptual learning, which is essential for scoring higher marks in school and competitive examinations.
Our Class 11 Mathematics NCERT Solutions cover all textbook questions, important exercise questions, and chapter-wise explanations in English Medium. Every solution is written in easy-to-understand language, allowing students to revise the chapter quickly before examinations. Regular practice of these solutions improves confidence, strengthens subject knowledge, and reduces examination stress.
Students preparing for school assessments should carefully study 11. Conic Sections because questions from this chapter are frequently asked in objective questions, short answer questions, long answer questions, competency-based questions, and case-study questions. Understanding the concepts explained in this chapter also helps students connect related topics from other chapters, making overall learning more effective and meaningful.
At ATP Education, we continuously update our Class 11 Mathematics English NCERT Solutions according to the latest NCERT textbooks and CBSE curriculum. Students can confidently use these chapter-wise solutions for daily study, homework assistance, quick revision, examination preparation, and self-learning. By studying 11. Conic Sections thoroughly and practising every question regularly, students can strengthen their concepts, improve writing skills, and achieve better academic performance in both school and board examinations.
11. Conic Sections - Class 11 Mathematics English NCERT Solutions
11. Conic Sections
Exercise 11.1 (Conic Sections)
Q1. Find the equation of the circle with centre (0, 2) and radius 2.
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (0, 2) and radius (r) = 2.
Therefore, the equation of the circle is
(x – 0)2 + (y – 2)2 = 22
⟹ x2 + y2 + 4 – 4 y = 4
⟹ x2 + y2 – 4y = 0
Q2. Find the equation of the circle with centre (–2, 3) and radius 4.
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–2, 3) and radius (r) = 4.
Therefore, the equation of the circle is (x + 2)2 + (y – 3)2 = (4)2
⟹ x2 + 4x + 4 + y2 – 6y + 9 = 16
⟹ x2 + y2 + 4x – 6y – 3 = 0
Q3.
Q10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 …………………. (1)
(6 – h)2 + (5 – k)2 = r2 …………………. (2)
Since the centre (h, k) of the circle lies on line 4x + y = 16,
∴ 4h + k = 16 …………………………………… (3)
From equations (1) and (2), we obtain
(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2
⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 ………………………………… (4)
On solving equations (3) and (4), we obtain h = 3 and k = 4.
On substituting the values of h and k in equation (1), we obtain
(4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
⇒ 𝑟=√10
Thus, the equation of the required circle is
(x – 3)2 + (y – 4)2 = (√10)2
x2 – 6x + 9 + y2 – 8y + 16 = 10
x2 + y2 – 6x – 8y + 15 = 0
The required equation for given circle is x2 + y2 – 6x – 8y + 15 = 0
11. Conic Sections
Exercise 11.2 (Conic Sections)
Q1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x.
Solution:
The given equation is y2 = 12x.
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y2 = 4ax, we obtain
4a = 12 ⇒ a = 3
∴ Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y2, the axis of the parabola is the x-axis.
Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12
Q3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = – 8x.
Solution:
The given equation is y2 = –8x.
Here, the coefficient of x is negative. Hence, the parabola opens towards the left.
On comparing this equation with y2 = –4ax, we obtain
–4a = –8 ⇒ a = 2
∴Coordinates of the focus = (–a, 0) = (–2, 0)
Since the given equation involves y2, the axis of the parabola is the x-axis.
Equation of directrix, x = a i.e., x = 2
Length of latus rectum = 4a = 8
Q4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = – 16y.
Solution:
The given equation is x2 = –16y.
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x2 = – 4ay, we obtain
–4a = –16 ⇒ a = 4
∴Coordinates of the focus = (0, –a) = (0, –4)
Since the given equation involves x2, the axis of the parabola is the y-axis.
Equation of directrix, y = a i.e., y = 4
Length of latus rectum = 4a = 16
Q7. Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6.
Solution:
Focus (6, 0); directrix, x = –6
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y2 = 4ax or
y2 = – 4ax.
It is also seen that the directrix, x = – 6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.
Hence, the parabola is of the form y2 = 4ax.
Here, a = 6
Thus, the equation of the parabola is y2 = 24x.
11. Conic Sections
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