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Solutions ⇒ Class 11th ⇒ Mathematics ⇒ 11. Conic Sections

# Exercise 11.2 Class 11 maths 11. Conic Sections - ncert solutions - Toppers Study

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## Exercise 11.2 Class 11 maths 11. Conic Sections - ncert solutions - Toppers Study

CBSE board students who preparing for class 11 ncert solutions maths and Mathematics solved exercise chapter 11. Conic Sections available and this helps in upcoming exams 2022-2023.

### You can Find Mathematics solution Class 11 Chapter 11. Conic Sections

• All Chapter review quick revision notes for chapter 11. Conic Sections Class 11
• NCERT Solutions And Textual questions Answers
• Extra NCERT Book questions Answers
• Importatnt key points with additional Assignment and questions bank solved.

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See all solutions for class 11 maths chapter 11 exercise 11 in english medium solved questions with answers.

## 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.