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# NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections Miscellaneous Exercise

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise - Free PDF Download

Chapter 10 Conic Sections covers the equations and characteristics of various conic sections, such as circles, ellipses, parabolas, and hyperbolas. These concepts are helpful for both advanced mathematics and solving geometry problems.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 10: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 10 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

It is important to practise the Miscellaneous Exercises from Chapter 10 to perform well on competitive tests and the CBSE Board exams as well. To practise offline, you can download the Class 11 Maths NCERT Solutions in PDF format. By completing such exercises, you will improve your knowledge and utilisation of conic section principles, which will help you be well-prepared for your tests. Get the latest CBSE Class 11 Maths Syllabus here.

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## Access NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

### Miscellaneous Exercise

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

Ans: As we know the origin of the coordinate plane is taken at the vertex of the parabolic reflector, where the axis of the reflector is along the positive x-axis.

The Diagrammatic representation is represented as follows:

As we know the equation of the parabola is of the form of ${{y}^{2}}=4ax$ (as it is opening to the right)

Since, the parabola passes through point $A \left( 5,10 \right),$

\begin{align} & \Rightarrow {{y}^{2}}=4ax \\ & \Rightarrow {{10}^{2}}=4\times a\times 5 \\ & \Rightarrow 100=20a \\ & \Rightarrow a=\dfrac{100}{20} \\ & \Rightarrow a=5 \\ \end{align}

The focus of the parabola is $\left( a,0 \right)=\left( 5,0 \right),$ which is the midpoint of the diameter.

Hence, the focus of the reflector is at the midpoint of the diameter.

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?

Ans: As we know the origin of the coordinate plane is taken at the vertex of the arch in such a way that its axis is along the y-axis.

The diagrammatic representation will be as follows:

The equation of the parabola is of the form ${{x}^{2}}=-4ay$ (as it is opening downwards).

Since the parabola passes through point $\left( \dfrac{5}{2},-10 \right),$

\begin{align} & {{\left( \dfrac{5}{2} \right)}^{2}}=-4\times a\times (-10) \\ & \Rightarrow a=\dfrac{25}{4\times 4\times 10}=\dfrac{5}{32} \\ \end{align}

Therefore,

The arch is in the form of a parabola whose equation is ${{x}^{2}}=-\dfrac{5}{8}y$

When $y=-2m,{{x}^{2}}=-\dfrac{5}{8}\times (-2)$ because it's 2m away downward from the vertex, so it will be negative.

\begin{align} & \Rightarrow {{x}^{2}}=\dfrac{5}{4} \\ & \Rightarrow x=\sqrt{\dfrac{5}{4}}m \\ \end{align}

The width of the parabola at 2m away from the vertex will be $=2\times \sqrt{\dfrac{5}{4}}m$

$=2\times 1.118m$(approx.)

$=2.23m$(approx.)

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23m.

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.

Ans: The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis.

This can be diagrammatically represented as

Here, AB and OC are the longest and shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadway, 18m from the middle.

Here, AB$=30m, OC=6m, and\; BC=\dfrac{100}{2}=50m$

The equation of the parabola is of the form ${{x}^{2}}=4ay$ (as it is opening upwards).

The coordinates of point A are $(50,30-6)=(50,24)$.

Since A$(50,24)$ is a point on the parabola,

\begin{align} & {{(50)}^{2}}=4a(24) \\ & \Rightarrow a=\dfrac{50\times 50}{4\times 24}=\dfrac{625}{24} \\ \end{align}

$\therefore$Equation of the parabola, ${{x}^{2}}=4\times \dfrac{625}{24}\times y$ or $6{{x}^{2}}=625y$

The x-coordinate of point D is 18.

Hence, at x = 18,

\begin{align} & 6{{(18)}^{2}}=625y \\ & \Rightarrow y=\dfrac{6\times 18\times 18}{625} \\ & \Rightarrow y=3.11(approx.) \\ \end{align}

$\therefore$DE = $3.11$m

DF = DE + EF = $3.11$ m + $6$ m = $9.11$ m

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately $9.11$ m.

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.

Ans: Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis.

Hence, the semi-ellipse can be diagrammatically represented as

The equation of the semi-ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1,y\ge 0,$ where a is the semi-major axis

Accordingly,

\begin{align} & 2a=8\Rightarrow a=4 \\ & b=2 \\ \end{align}

Therefore, the equation of the semi-ellipse is $\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1,y\ge 0,$ ….(1)

Let A be a point on the major axis such that AB = 1.5 m.

Draw AC ⊥ OB.

OA = (4 – 1.5) m = 2.5 m

The x-coordinate of point C is 2.5.

On substituting the value of x with 2.5 in equation (1), we’ll get

\begin{align} & \dfrac{{{\left( 2.5 \right)}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1 \\ & \Rightarrow \dfrac{6.25}{16}+\dfrac{{{y}^{2}}}{4}=1 \\ & \Rightarrow {{y}^{2}}=4\left( 1-\dfrac{6.25}{16} \right) \\ & \Rightarrow {{y}^{2}}=4\left( \dfrac{9.75}{16} \right) \\ & \Rightarrow {{y}^{2}}=2.4375 \\ & \Rightarrow {y}=1.56(approx.) \\ & \therefore AC=1.56m \\ \end{align}

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

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.

Ans: Let AB be the rod making an angle θ with OX and P (x, y) be the point on it such that AP = 3 cm.

Then, PB = AB – AP = (12 – 3) cm = 9 cm [AB = 12 cm]

From P, draw PQ⊥OY and PR⊥OX.

In ΔPBQ, $\cos \theta =\dfrac{PQ}{PB}=\dfrac{x}{9}$

In ΔPRA, $\sin \theta =\dfrac{PR}{PA}=\dfrac{y}{3}$

Since, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$

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

Or,  ${{\dfrac{x}{81}}^{2}}+{{\dfrac{y}{9}}^{2}}=1$

Thus, the equation of the locus of point P on the rod is ${{\dfrac{x}{81}}^{2}}+{{\dfrac{y}{9}}^{2}}=1$

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.

Ans: The given parabola is ${{x}^{2}}=12y$.

On comparing this equation with ${{x}^{2}}=4ay$, we’ll get $4a=12\Rightarrow a=3$

$\therefore$The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At $y=3,$ ${{x}^{2}}=12(3)\Rightarrow {{x}^{2}}=36\Rightarrow x=\pm 6$

$\therefore$The coordinates of A are (–6, 3), while the coordinates of B are (6, 3).

Therefore, the vertices of ΔOAB are O (0, 0), A (–6, 3), and B (6, 3).

Area of ΔOAB $=\dfrac{1}{2}\left| 0\left( 3-3 \right)+\left( -6 \right)\left( 3-0 \right)+6\left( 0-3 \right) \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\left| \left( -6 \right)\left( 3 \right)+6\left( -3 \right) \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\left| -18-18 \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\left| -36 \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\times 36uni{{t}^{2}}$

$=18$ $uni{{t}^{2}}$

Thus, the required area of the triangle is $18$ $uni{{t}^{2}}$.

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.

Ans: Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form of $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1,$ where a is the semi-major axis.

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

Distance between the foci $(2c)=8\Rightarrow c=4$

On using the relation $c=\sqrt{{{a}^{2}}-{{b}^{2}}},$ we’ll get

$4=\sqrt{25-{{b}^{2}}}$

$\Rightarrow 16=25-{{b}^{2}}$

$\Rightarrow {{b}^{2}}=25-16$

$\Rightarrow {{b}^{2}}=9$

$\Rightarrow b=3$

Thus, the equation of the path traced by the man is $\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1$

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

Ans: Let OAB be the equilateral triangle inscribed in parabola ${{y}^{2}}=4ax$

Let AB intersect the x-axis at point C.

Let OC = k

From the equation of the given parabola, we have ${{y}^{2}}=4ak\Rightarrow y=\pm 2\sqrt{ak}$

$\therefore$The respective coordinates of points A and B are $\left( k,-2\sqrt{ak} \right)$ and $\left( k,-2\sqrt{ak} \right)$

AB = CA + CB = $2\sqrt{ak}+2\sqrt{ak}=4\sqrt{ak}$

Since OAB is an equilateral triangle, $O{{A}^{2}}=A{{B}^{2}}.$

$\therefore {{k}^{2}}+{{\left( 2\sqrt{ak} \right)}^{2}}={{\left( 4\sqrt{ak} \right)}^{2}}$

$\Rightarrow {{k}^{2}}+4ak=16ak$

$\Rightarrow {{k}^{2}}=12ak$

$\Rightarrow k=12a$

$\therefore AB=4\sqrt{ak}=4\sqrt{a\times 12a}$

$=4\sqrt{12{{a}^{2}}}$

$=8\sqrt{3}a$

Thus, the side of the equilateral triangle inscribed in parabola ${{y}^{2}}=4ax$ is $8\sqrt{3}a$

## Conclusion

NCERT Miscellaneous Exercise in Class 11 Maths Chapter 10 is important for learning about conic sections. It includes many problems that help you understand circles, ellipses, parabolas, and hyperbolas better. Focus on solving these problems to get better at finding equations and answering geometry questions. Regular practice will make you more confident and ready for exams. Make sure to go through the solutions carefully to understand the concepts well.

## Class 11 Maths Chapter 10: Exercises Breakdown

 Exercise Number of Questions Exercise 10.1 15 Questions & Solutions Exercise 10.2 12 Questions & Solutions Exercise 10.3 20 Questions & Solutions Exercise 10.4 15 Questions & Solutions

## CBSE Class 11 Maths Chapter 10 Other Study Materials

 S.No Important Links for Chapter 10 Conic Sections 1 Class 11 Conic Sections Revision Notes 2 Class 11 Conic Sections Important Questions 3 Class 11 Conic Sections NCERT Exemplar Solution

## 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.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections Miscellaneous Exercise

1. What is the focus of the NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 10?

In NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 10 reviews conic sections, including circles, parabolas, ellipses, and hyperbolas. It integrates various concepts learned throughout the Chapter. The exercise ensures a thorough understanding and application of these concepts.

2. Why are conic sections important in NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 10?

In NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 10, conic sections are crucial in mathematics due to their presence in real-life applications like satellite orbits, architectural designs, and physics. Understanding these shapes aids in solving complex problems related to these areas, making them essential for students to master.

3. How can one identify the type of conic section from its equation in NCERT Solutions of Conic Sections Class 11 Miscellaneous Exercise?

Identifying conic sections involves analysing the general equation $Ax^{2}+Byx+Cy^{2}+Dx+Ey+F=0$. The coefficients and discriminant determine if it is a circle, ellipse, parabola, or hyperbola. Understanding these criteria is essential for solving related problems.

4. What are key strategies for solving problems in NCERT Solutions of Conic Sections Class 11 Miscellaneous Exercise?

Key strategies include understanding the standard forms of conic sections and practising completing the square for equations. Applying geometric properties and drawing diagrams helps visualize and solve the problems effectively, making these strategies crucial for success.

5. Are there any real-life applications included in NCERT Solutions of Conic Sections Class 11 Miscellaneous Exercise?

Yes, Conic Sections Class 11 Miscellaneous Exercise some problems involve real-life scenarios such as planetary paths, reflective surface designs, and trajectories in physics. These applications illustrate the practical significance of conic sections, helping students understand their relevance beyond textbook problems.

6. What should students focus on while preparing for this NCERT Solution of Class 11 Maths Conic Sections Miscellaneous Exercise?

Students should focus on mastering the derivation and properties of conic sections. Practising a variety of problems and understanding how to transition between different forms of equations are essential. Consistent practice and reviewing solved examples are crucial for success.

7. What topics are covered in the Class 11 Maths Conic Sections Miscellaneous Exercise?

The Miscellaneous Exercise covers various topics related to conic sections, including the equations and properties of circles, ellipses, parabolas, and hyperbolas.

8. How do I solve problems involving the equation of a circle in Class 11 Maths Conic Sections Miscellaneous Exercise?

To solve these problems from Class 11 Maths Conic Sections Miscellaneous Exercise, you need to understand how to derive the equation of a circle given its centre and radius, and how to apply this equation to different scenarios.

9. What kind of questions involve ellipses in the Class 11 Maths Chapter 10 Miscellaneous Solutions?

Questions involving ellipses may require you to find the standard form of the ellipse equation, identify its major and minor axes, and solve problems related to its geometric properties.

10. How are parabolas addressed in the Class 11 Maths Chapter 10 Miscellaneous Solutions?

Problems related to parabolas might include finding their equations, determining the focus and directrix, and solving real-world applications using these properties.

11. What is the approach to solving hyperbola-related questions in Class 11 Maths Chapter 10 Miscellaneous Solutions?

Class 11 Maths Chapter 10 Miscellaneous Solutions for hyperbolas, you need to know how to derive their equations, understand their asymptotes, and solve problems involving their geometric features.

12. Why is practising the Class 11 Maths Chapter 10 Miscellaneous Solutions important for exams?

Practising Class 11 Maths Chapter 10 Miscellaneous Solutions helps you thoroughly understand the concepts of conic sections, improves your problem-solving skills, and ensures you are well-prepared for board exams and competitive tests.