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11. Conic Sections MATHEMATICS class 11 in English Medium ncert book solutions Exercise 11.2
11. Conic Sections Exercise 11.2 – Complete NCERT Book Solutions for Class 11 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 11. Conic Sections Exercise 11.2 to help you master concepts and score higher.
11. Conic Sections MATHEMATICS class 11 in English Medium ncert book solutions Exercise 11.2
NCERT Solutions for Class 11 Mathematics play an important role in helping students understand the concepts of the chapter 11. Conic Sections clearly. This chapter includes the topic Exercise 11.2, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 11 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 11. Conic Sections is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 11.2. By studying these updated NCERT Solutions for Class 11 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
11. Conic Sections
Exercise 11.2
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.
See other sub-topics of this chapter:
1. Exercise 11.1 2. Exercise 11.2 3. Exercise 11.3 4. Exercise 11.4 5. Miscellaneous Exercise on Chapter - 11