NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

Edited By Vishal kumar | Updated on Nov 20, 2023 11:49 AM IST

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 11: Conic Sections Miscellaneous Exercise- In the previous exercises of this chapter, you have already learned about the conic sections such as circle, parabola, hyperbola, and ellipse. In the NCERT solutions for Class 11 Maths chapter 11 miscellaneous exercise, you will get mixed concepts questions from all the topics of the conic sections. Miscellaneous exercise Chapter 11 Class 11 is considered to be tougher than all the previous exercises of this chapter. It will check your conceptual understanding of this chapter.

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise- Download Free PDF
NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Miscellaneous Exercise
Access Conic Section Class 11 Chapter 11 Miscellaneous Exercise
More About NCERT Solutions for Class 11 Maths Chapter 11 Miscellaneous Exercise:-
Topic Covered in Class 11 Maths ch 11 Miscellaneous Exercise Solutions
Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Miscellaneous Exercise:-
Key Features of Class 11 Chapter 11 Maths Miscellaneous Solutions
NCERT Solutions of Class 11 Subject Wise
Subject Wise NCERT Exampler Solutions

If you are done with the other exercises of this chapter, you start solving problems from miscellaneous exercise chapter 11 Class 11. You will find difficulties while solving these problems at first but Class 11 Maths chapter 11 miscellaneous solutions are here to help you. These NCERT solutions are designed in a detailed manner that could be understood by an average student also. There are very few questions in the CBSE Exam that are asked for the miscellaneous exercise, but miscellaneous exercises are very important for the engineering competitive exams like JEE Main, SRMJEE, etc. You must solve the miscellaneous exercises if you are preparing for any engineering entrance exams. The class 11 chapter 11 maths miscellaneous solutions are written by subject experts from Careers360, presented in easily understandable language with detailed explanations. Also, the class 11 maths ch 11 miscellaneous exercise solutions are available in a downloadable PDF format, allowing students to access and use them anytime, free of charge. This resource is designed to enhance students' understanding and facilitate convenient and cost-free self-study. If you are looking for NCERT solutions for other classes, click on the NCERT Solutions link.

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**In the CBSE Syllabus for 2023-24, this miscellaneous exercise class 11 chapter 11 has been renumbered as Chapter 10.

NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Miscellaneous Exercise

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Access Conic Section Class 11 Chapter 11 Miscellaneous Exercise

Question:1 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Answer:

Le the parabolic reflector opens towards the right.

So the equation of parabolic reflector will be,

$y^2=4ax$

Now, Since this curve will pass through the point (5,10) if we assume origin at the optical centre,

$10^2=4a(5)$

$a=\frac{100}{20}=5$

Hence, The focus of the parabola is,

$(a,0)=(5,0)$ .

Alternative Method,

As we know on any concave curve

$f=\frac{R}{2}$

Hence, Focus

$f=\frac{R}{2}=\frac{10}{2}=5$ .

Hence the focus is 5 cm right to the optical centre.

Question:2 An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Answer:

Since the Axis of the parabola is vertical, Let the equation of the parabola be,

$x^2=4ay$

it can be seen that this curve will pass through the point (5/2, 10) if we assume origin at the bottom end of the parabolic arch.

So,

$\left(\frac{5}{2}\right)^2=4a(10)$

$a=\frac{25}{160}=\frac{5}{32}$

Hence, the equation of the parabola is

$x^2=4\times\frac{5}{32}\times y$

$x^2=\frac{20}{32} y$

$x^2=\frac{5}{8} y$

Now, when y = 2 the value of x will be

$x=\sqrt{(\frac{5}{8}\times2)}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$

Hence the width of the arch at this height is

$2x=2\times\frac{\sqrt{5}}{2}=\sqrt{5}.$

Question:3 The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m Find the length of a supporting wire attached to the roadway 18 m from the middle

Answer:

Given,

The width of the parabolic cable = 100m

The length of the shorter supportive wire attached = 6m

The length of the longer supportive wire attached = 30m

Since the rope opens towards upwards, the equation will be of the form

$x^2=4ay$

Now if we consider origin at the centre of the rope, the equation of the curve will pass through points, (50,30-6)=(50,24)

$24^2=4a50$

$a=\frac{625}{24}$

Hence the equation of the parabola is

$x^2=4\times \frac{625}{24}\times y$

$x^2= \frac{625}{6}\times y$

Now at a point, 18 m right from the centre of the rope, the x coordinate of that point will be 18, so by the equation, the y-coordinate will be

$y=\frac{x^2}{4a}=\frac{18^2}{4\times \frac{625}{6}}\approx 3.11m$

Hence the length of the supporting wire attached to roadway from the middle is 3.11+6=9.11m.

Question:4 An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Answer:

The equation of the semi-ellipse will be of the form

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\:,y>0$

Now, According to the question,

the length of major axis = 2a = 8 $\Rightarrow a=4$

The length of the semimajor axis =2 $\Rightarrow b=2$

Hence the equation will be,

$\frac{x^2}{4^2}+\frac{y^2}{2^2}=1\:,y>0$

$\frac{x^2}{16}+\frac{y^2}{4}=1\:,y>0$

Now, at point 1.5 cm from the end, the x coordinate is 4-1.5 = 2.5

So, the height at this point is

$\frac{(2.5)^2}{16}+\frac{y^2}{4}=1\Rightarrow y=\sqrt{4(1-\frac{2.5^2}{16})}$

$y\approx 1.56m$

Hence the height of the required point is 1.56 m.

Question:5 A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Answer:

Let $\theta$ be the angle that rod makes with the ground,

Now, at a point 3 cm from the end,

$\cos\theta=\frac{x}{9}$

At the point touching the ground

$\sin\theta=\frac{y}{3}$

Now, As we know the trigonometric identity,

$\sin^2\theta+\cos^2\theta=1$

$\left (\frac{x}{9} \right )^2+\left ( \frac{y}{3} \right )^2=1$

$\frac{x^2}{81}+\frac{y^2}{9}=1$

Hence the equation is,

$\frac{x^2}{81}+\frac{y^2}{9}=1$

Question:6 Find the area of the triangle formed by the lines joining the vertex of the parabola $x ^2 = 12y$ to the ends of its latus rectum.

Answer:

Given the parabola,

$x^2=12y$

Comparing this equation with $x^2=4ay$ , we get

$a=3$

Now, As we know the coordinates of ends of latus rectum are:

$(2a,a)\:and\:(-2a,a)$

So, the coordinates of latus rectum are,

$(2a,a)\:and\:(-2a,a)=(6,3)\:and\:(-6,3)$

Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3)

Widht of the triangle = 2*6=12

Height of the triangle = 3

So The area =

$\frac{1}{2}\times base\times height=\frac{1}{2}\times12\times3=18$

Hence the required area is 18 unit square.

Question:7 A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Answer:

As we know that if a point moves in a plane in such a way that its distance from two-point remain constant then the path is an ellipse.

Now, According to the question,

the distance between the point from where the sum of the distance from a point is constant = 10

$\Rightarrow 2a=10\Rightarrow a=5$

Now, the distance between the foci=8

$\Rightarrow 2c=8\Rightarrow c=4$

Now, As we know the relation,

$c^2=a^2-b^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3$

Hence the equation of the ellipse is,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$

Hence the path of the man will be

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$

Question:8 An equilateral triangle is inscribed in the parabola $y^2 = 4 ax$ , where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer:

Given, an equilateral triangle inscribed in parabola with the equation. $y^2 = 4 ax$

The one coordinate of the triangle is A(0,0).

Now, let the other two coordinates of the triangle are

$B(x,\sqrt{4ax})$ and $C(x,-\sqrt{4ax})$

Now, Since the triangle is equilateral,

$BC=AB=CA$

$2\sqrt{4ax}=\sqrt{(x-0)^2+(\sqrt{4ax}-0)^2}$

$x^2=12ax$

$x=12a$

The coordinates of the points of the equilateral triangle are,

$(0,0),(12,\sqrt{4a\times 12a}),(12,-\sqrt{4a\times 12a})=(0,0),(12,4\sqrt{3}a)\:and\:(12,-4\sqrt{3}a)$

So, the side of the triangle is

$2\sqrt{4ax}=2\times4\sqrt{3}a=8\sqrt{3}a$

Topic Covered in Class 11 Maths ch 11 Miscellaneous Exercise Solutions

The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 11 - Conic Sections is centred around the following topics:

Sections of a Cone
Circle
Parabola
Ellipse
Hyperbola

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Each question in the class 11 chapter 11 miscellaneous exercise has been meticulously solved to aid students in understanding the concepts, with an eye on the examination. The NCERT Solutions for Class 11 Maths play a crucial role in providing in-depth comprehension of all the concepts covered in Class 11 Mathematics.

Also Read| Conic Section Class 11 Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Miscellaneous Exercise:-

Class 11 Maths chapter 11 miscellaneous exercise solutions are explained in a detailed manner by experts who knows how to answer the in the CBSE exam.
You can use NCERT solutions for Class 11 Maths chapter 11 miscellaneous exercise for reference while solving them.

Key Features of Class 11 Chapter 11 Maths Miscellaneous Solutions

Key features of Class 11 Maths Chapter 11 Miscellaneous Exercise Solutions on Conic Sections include:

Comprehensive Coverage: Miscellaneous exercise class 11 chapter 11 solution covers all the miscellaneous exercises in Chapter 11, addressing key topics related to Conic Sections.
Careful Problem Solving: Each class 11 maths miscellaneous exercise chapter 11 question is carefully solved, ensuring clarity and understanding for students.
Exam-Oriented Approach: Class 11 maths ch 11 miscellaneous exercise solutions are crafted with the examination in mind, providing students with a strategic approach to problem-solving.
Variety of Problems: A diverse range of problems is included in the miscellaneous exercise, catering to different difficulty levels and scenarios.
Accessible Language: Class 11 chapter 11 miscellaneous exercise solutions are presented in clear and concise language, making complex geometric concepts associated with Conic Sections more accessible for students.
PDF Version Access: The solutions are available in a downloadable PDF format, offering students convenient offline access for study.

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NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

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Frequently Asked Questions (FAQs)

1. Write the equation of a circle with centre (1,2) and the radius 5 ?

Equation of circle => (x-1)^2 + (y-2)^2 = 5^2

x^2 -2x + 1 + y^2 + 4y + 4 = 25

x^2 -2x + y^2 + 4y = 20

2. Write the equation of the parabola with focus at (2, 0) and directrix x = – 2 ?

Equation of the parabola => y^2 = 4(2)x

y^2 = 8x

3. Find the coordinates of the focus of the parabola y^2 = 12x

Given equation of the parabola => y^2 = 12x

Compare with y^2 = 4ax => a = 3

Coordinate of focus is (3,0).

4. Find the axis of the parabola y^2 = 12x

The given equation involves y^2, so the axis of symmetry is along the x-axis. The axis of the parabola is the x-axis.

5. Find the equation of the directrix of the parabola y^2 = 12x

Given equation of the parabola => y^2 = 12x

Compare with y^2 = 4ax => a = 3

The equation of the directrix of the parabola is x = – 3

6. Find the the length of the latus rectum of the parabola y^2 = 12x

Given equation of the parabola => y^2 = 12x

Compare with y^2 = 4ax => a = 3

Length of the latus rectum is 4a = 4 × 3 = 12.

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