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Download FREE PDF for NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections: Exercise 10.2
Download the FREE PDF of NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections: Exercise 10.2. This exercise covers detailed solutions related to parabolas and their properties, including how to derive the equation, locate the focus, directrix, and axis of symmetry. The Class 11 Maths NCERT Solutions which are key parts of the Class 11 Maths Syllabus and designed to help you solve problems easily while strengthening your understanding of parabolas.
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To access these solutions, click on the link below. This PDF will not only help you with your homework but also serve as a quick revision guide before exams. Make sure to download it and keep it handy!
Glance on NCERT Solutions Exercise 10.2 of Class 11 Maths - Conic Sections
Focuses on the parabola and its key characteristics: Exercise 10.2 is dedicated to understanding the structure of a parabola, including important elements like the vertex (the point where the parabola changes direction), the focus (a fixed point inside the curve), and the directrix (a fixed line). These characteristics are essential for defining the shape and orientation of the parabola.
Introduces the standard equation of a parabola: The exercise covers the standard equations of parabolas in different orientations, horizontal and vertical. Students learn how these equations relate to the position of the focus and directrix, making it easier to identify and work with parabolas in various problems.
Explains how to find the axis of symmetry and the position of the parabola: The NCERT Solutions provide guidance on finding the axis of symmetry, which is the line that divides the parabola into two equal halves. This axis passes through the vertex and is key to understanding the parabola’s orientation.
Guides students on how to derive the equation of a parabola: This exercise teaches how to derive the equation of a parabola based on given conditions, such as the location of the focus and the directrix. These derivations help students gain a deeper understanding of how a parabola’s equation reflects its geometric properties.
Provides step-by-step solutions for solving problems related to parabolic curves: The NCERT Solutions offer detailed explanations for each question in the exercise, breaking down the process into clear, manageable steps. This helps students solve problems more effectively and reinforces their understanding of the concepts.
Essential for understanding the geometric and algebraic properties of parabolas: Mastering this exercise is crucial for students as parabolas play a significant role in advanced geometry and calculus. The knowledge gained here will be helpful in solving real-world problems and understanding higher-level mathematical concepts.
Formulas Used in Class 11 Maths Exercise 10.2
In Exercise 10.2 of Class 11 Maths, which focuses on parabolas, several key formulas are used:
1. Standard Equation of a Parabola (Horizontal Axis):
- $y^2 = 4ax$, where $a$ is the distance from the vertex to the focus.
2. Standard Equation of a Parabola (Vertical Axis):
- $x^2 = 4ay$, where $a$ is the distance from the vertex to the focus.
3. Vertex of a Parabola:
- The vertex is the point $(0, 0)$ in the standard form, unless shifted by translation.
4. Focus of a Parabola:
- For the parabola $y^2 = 4ax$, the focus is at $(a, 0)$.
- For the parabola $x^2 = 4ay$, the focus is at $(0, a)$.
5. Equation of the Directrix:
- For $y^2 = 4ax$, the directrix is $x = -a$.
- For $x^2 = 4ay$, the directrix is $y = -a$.
6. Axis of Symmetry:
- For $y^2 = 4ax$, the axis of symmetry is the x-axis.
- For $x^2 = 4ay$, the axis of symmetry is the y-axis.
NCERT Solutions Class 11 Maths Chapter 10 Conic Sections Exercise 10.2
ShareShare1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${{y}^{2}}=12x$
Ans: The given equation is ${{y}^{2}}=12x$.
Here, the coefficient of x is positive. Hence, the parabola opens towards the right. On comparing this equation with ${{y}^{2}}=4ax,$ we’ll get
$4a=12\Rightarrow a=3$
$\therefore $ Coordinates of the focus = $=(a,0)=(3,0)$
Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=-a\text{ }i.e.,\text{ }x=-3\text{ }i.e.,\text{ }x+3=0$
Length of latus rectum $=4a=4\times 3=12$
2. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${{x}^{2}}=6y$
Ans: The given equation is ${{x}^{2}}=6y$.
Here, the coefficient of y is positive. Hence, the parabola opens upwards.
On comparing this equation with ${{x}^{2}}=4ay$ we obtain
$4a=6\Rightarrow a=\dfrac{3}{2}$
$\therefore $Coordinates of the focus $=(0,a)=\left( 0,\dfrac{3}{2} \right)$
Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=-a$ i.e., $y=\dfrac{-3}{2}$
Length of latus rectum $=4a=6$
3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${{y}^{2}}=-8x$
Ans: The given equation is ${{y}^{2}}=-8x$.
Here, the coefficient of $x$ is negative. Hence, the parabola opens towards the left. On comparing this equation with ${{y}^{2}}=-4ax,$ we’ll get
$-4a=-8\Rightarrow a=2$
$\therefore $ Coordinates of the focus $=(-a,0)=(-2,0)$
Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=a$ i.e., $x=2$
Length of latus rectum $=4a=8$
4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${{x}^{2}}=-16y$
Ans: The given equation is ${{x}^{2}}=-16y.$
Here, the coefficient of $y$ is negative. Hence, the parabola opens downwards.
On comparing this equation with ${{x}^{2}}=-4ay$, we’ll get
$-4a=-16\Rightarrow a=4$
$\therefore $ Coordinates of the focus $=(0,-a)=(0,-4)$
Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=a$ i.e., $y=4$
Length of latus rectum $=4a=16$
5. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${{y}^{2}}=10x$
Ans:The given equation is ${{y}^{2}}=10x$.
Here, the coefficient of $x$ is positive. Hence, the parabola opens towards the right. On comparing this equation with ${{y}^{2}}=4ax$, we’ll get
$4a=10\Rightarrow a=\dfrac{5}{2}$
$\therefore $ Coordinates of the focus $=(a,0)=\left( \dfrac{5}{2},0 \right)$
Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=-a$ i.e., $x=-\dfrac{5}{2}$
Length of latus rectum $=4a=10$
6. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${{x}^{2}}=-9y$
Ans: The given equation is ${{x}^{2}}=-9y$.
Here, the coefficient of $y$ is negative. Hence, the parabola opens downwards.
On comparing this equation with ${{x}^{2}}=-4ay$, we’ll get
$-4a=-9\Rightarrow a=\dfrac{9}{4}$
$\therefore $ Coordinates of the focus $=(0,-a)=\left( 0,-\dfrac{9}{4} \right)$
Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=a$ i.e., $y=\dfrac{9}{4}$
Length of latus rectum $=4a=9$
7. Find the equation of the parabola that satisfies the following conditions: Focus $(6,0);$ directrix $x=-6$
Ans: 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 ${{y}^{2}}=4ax$ or ${{y}^{2}}=-4ax$.
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 ${{y}^{2}}=4ax$.
Here, $a=6$
Thus, the equation of the parabola is ${{y}^{2}}=24x$.
8. Find the equation of the parabola that satisfies the following conditions: Focus $(0,-3);$ directrix $y=3$
Ans: Focus $=(0,-3);$ directrix $y=3$
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form ${{x}^{2}}=4ay$ or ${{x}^{2}}=-4ay.$
It is also seen that the directrix, $y=3$ is above the x-axis, while the focus $(0,-3)$ is below the x-axis.
Hence, the parabola is of the form ${{x}^{2}}=-4ay.$
Here, $a=3$
Thus, the equation of the parabola is ${{x}^{2}}=-12y.$
9. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0);$ focus $(3,0)$
Ans: Vertex $(0,0);$ focus $(3,0)$
Since the vertex of the parabola is $(0,0)$ and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form ${{y}^{2}}=4ax.$
Since the focus is $(3,0)$, $a=3$.
Thus, the equation of the parabola is ${{y}^{2}}=4\times 3\times x$ i.e., ${{y}^{2}}=12x$
10. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0)$ focus $(-2,0)$
Ans: Solution 10: Vertex $(0,0)$ focus $(-2,0)$
Since the vertex of the parabola is $(0,0)$ and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form ${{y}^{2}}=-4ax.$
Since the focus is $(-2,0),$$a=2.$
Thus, the equation of the parabola is ${{y}^{2}}=-4\times 2\times x$ i.e., ${{y}^{2}}=-8x$
11. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0)$passing through $(2,3)$and axis is along x-axis
Ans: Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form ${{y}^{2}}=4ax$ or ${{y}^{2}}=-4ax.$
The parabola passes through point $(2,3)$, which lies in the first quadrant. Therefore, the equation of the parabola is of the form ${{y}^{2}}=4ax$, while point $(2,3)$ must satisfy the equation ${{y}^{2}}=4ax$.
$\therefore {{(3)}^{2}}=4a(2)\Rightarrow a=\dfrac{9}{8}$
Thus, the equation of the parabola is ${{y}^{2}}=4\left( \dfrac{9}{8} \right)x$
$\begin{align} & \Rightarrow {{y}^{2}}=\dfrac{9}{2}x \\ & \Rightarrow 2{{y}^{2}}=9x \\ \end{align}$
12. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0)$, passing through $(5,2)$ and symmetric with respect to y-axis
Ans: Since the vertex is $(0,0)$ and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form ${{x}^{2}}=4ay$ or ${{x}^{2}}=-4ay.$
The parabola passes through point $(5,2)$, which lies in the first quadrant. Therefore, the equation of the parabola is of the form ${{x}^{2}}=4ay$, while point $(5,2)$ must satisfy the equation ${{x}^{2}}=4ay$.
$\begin{align} & \therefore {{(5)}^{2}}=4\times a\times 2 \\ & \Rightarrow 25=8a \\ & \Rightarrow a=\dfrac{25}{8} \\ \end{align}$
Thus, the equation of the parabola is
$\Rightarrow {{x}^{2}}=4\left( \dfrac{25}{8} \right)y$
$\Rightarrow 2{{x}^{2}}=25y$
Class 11 Maths Chapter 10: Exercises Breakdown
Conclusion
NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections: Exercise 10.2 provide a detailed understanding of parabolas and their properties, including the focus, directrix, and axis of symmetry. These step-by-step solutions help simplify the complex equations of parabolas, making it easier for students to grasp the concepts effectively. By working through these solutions, students can build a solid foundation for more advanced geometry and calculus topics, and gain confidence in tackling exam questions related to conic sections. Make sure to download the FREE PDF for easy access and effective preparation.
CBSE Class 11 Maths Chapter 10 - Conic Sections Other Study Materials
Chapter-Specific NCERT Solutions for Class 11 Maths
Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
Additional Study Materials for Class 11 Maths
FAQs on NCERT Solutions Class 11 Maths Chapter 10 Conic Sections Exercise 10.2
1: What is the main focus of Exercise 10.2 in Chapter 10 of Class 11 Maths?
Exercise 10.2 focuses on parabolas, covering their properties, such as the vertex, focus, directrix, and axis of symmetry. It provides practice in understanding the standard forms and how to derive the equation of a parabola.
2: How do NCERT Solutions for Exercise 10.2 help in understanding parabolas?
The solutions offer step-by-step explanations to solve problems involving parabolas. They make it easier to understand how different elements like the focus and directrix relate to the equation of a parabola.
3: Are NCERT Solutions for Exercise 10.2 available for free?
Yes, you can download the NCERT Solutions for Exercise 10.2 from Vedantu’s website for free in PDF format to help you study and revise effectively.
4: What is the importance of learning parabolas in conic sections?
Parabolas are an important type of conic section with practical applications in physics, engineering, and architecture. Understanding parabolas helps in solving problems in various branches of science and mathematics.
5: Which equations are covered in Exercise 10.2 for parabolas?
Exercise 10.2 includes the standard equations of parabolas, such as $y^2 = 4ax$ for horizontal parabolas and $x^2 = 4ay$ for vertical parabolas, and how to derive these based on given conditions.
6: How difficult is Exercise 10.2 for students?
Exercise 10.2 can be moderately challenging as it involves new concepts like focus, directrix, and deriving equations. However, the NCERT Solutions make these concepts easier to understand by breaking down each problem into manageable steps.
7: Do the NCERT Solutions for Exercise 10.2 include diagrams for better understanding?
Yes, the solutions include diagrams that help visualise parabolas, making it easier to understand their properties and how their equations are derived.
8: Can these NCERT Solutions help in competitive exams?
Yes, understanding parabolas is essential for competitive exams like JEE, where conic sections are frequently tested. The NCERT Solutions help in building the fundamentals needed for such exams.
9: Are parabolas only discussed in one orientation in Exercise 10.2?
No, Exercise 10.2 covers parabolas in both horizontal and vertical orientations, explaining how their focus, directrix, and axis of symmetry vary based on orientation.
10: How are the concepts of focus and directrix important in Exercise 10.2?
The focus and directrix are key elements that define the shape and position of a parabola. Understanding these helps in deriving the equation and graphing the parabola accurately.
11: What are some real-world applications of parabolas discussed in this exercise?
Though the exercise primarily focuses on mathematical aspects, parabolas have real-world applications like satellite dishes, car headlights, and other structures that utilise the reflective properties of parabolas.
12: How do I download the PDF for NCERT Solutions of Exercise 10.2?
You can download the PDF from the link provided on the page, allowing you to study offline at your convenience.
13: Are these NCERT Solutions enough for understanding the chapter thoroughly?
Yes, these NCERT Solutions provide comprehensive explanations for Exercise 10.2. For deeper understanding, it is also helpful to practise additional problems from reference books or previous exam papers.
