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NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Conic Section

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Conic Section

Edited By Komal Miglani | Updated on May 06, 2025 04:27 PM IST

From the perfect curve of a football’s path to the elliptical orbits of planets, conic sections are all around us. These curves- circle, ellipse, parabola, and hyperbola are formed by slicing a cone at different angles. In mathematics, they will help you to model motion, design structures and even focus light and sound! In this exercise, you will apply all the key concepts of conic sections, like equations of circles, ellipses, parabolas, hyperbolas, etc. The questions are a blend of geometry and algebra that will test your overall understanding of the chapter.

This Story also Contains
  1. Class 11 Maths Chapter 10 Conic Sections Miscellaneous Exercise Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 10: Miscellaneous Exercise
  3. Topics covered in Chapter 10 Conic Sections: Miscellaneous Exercise
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions

The NCERT is a key guide that offers step-by-step calculations and clear explanations to help you tackle a variety of problems. These NCERT solutions will help you improve your speed and accuracy and will prepare you for upcoming exams.

Class 11 Maths Chapter 10 Conic Sections Miscellaneous Exercise Solutions - Download PDF

Download PDF


NCERT Solutions Class 11 Maths Chapter 10: Miscellaneous Exercise

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

Answer:

The parabolic reflector opens towards the right.

So the equation of a parabolic reflector will be,

y2=4ax

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

So

102=4a(5)

a=10020=5

Hence, the focus of the parabola is,

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

Alternative Method,

As we know, on any concave curve

f=R2

Hence, Focus

f=R2=102=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,

x2=4ay

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

So,

(52)2=4a(10)

a=25160=532

Hence, the equation of the parabola is

x2=4×532×y

x2=2032y

x2=58y

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

x=(58×2)=54=52

Hence, the width of the arch at this height is

2x=2×52=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 upwards, the equation will be of the form

x2=4ay

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

242=4a50

a=62524

Hence, the equation of the parabola is

x2=4×62524×y

x2=6256×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=x24a=1824×62563.11m

Hence, the length of the supporting wire attached to the 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

x2a2+y2b2=1,y>0

Now, according to the question,

the length of major axis = 2a = 8 a=4

The length of the semimajor axis =2b=2

Hence, the equation will be,

x242+y222=1,y>0

x216+y24=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

(2.5)216+y24=1y=4(12.5216)

y1.56m

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 θ be the angle that rod makes with the ground,

Now, at a point 3 cm from the end,

cosθ=x9

At the point touching the ground

sinθ=y3

Now, as we know the trigonometric identity,

sin2θ+cos2θ=1

(x9)2+(y3)2=1

x281+y29=1

Hence, the equation is,

x281+y29=1

Question 6: Find the area of the triangle formed by the lines joining the vertex of the parabola x2=12y to the ends of its latus rectum.

Answer:

Given the parabola,

x2=12y

Comparing this equation with x2=4ay, we get

a=3

Now, as we know, the coordinates of the ends of the latus rectum are

(2a,a)and(2a,a)

So, the coordinates of the 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)

Width of the triangle = 2*6=12

Height of the triangle = 3

So the area =

12×base×height=12×12×3=18

Hence, the required area is 18 units 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, if a point moves in a plane in such a way that its distance from two points remains constant, then the path is an ellipse.

Now, according to the question,

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

2a=10a=5

Now, the distance between the foci=8

2c=8c=4

Now, as we know the relation,

c2=a2b2

b2=a2c2

b=a2c2=5242=2516=9=3

Hence, the equation of the ellipse is,

x2a2+y2b2=1

x252+y232=1

x225+y29=1

Hence, the path of the man will be

x225+y29=1

Question 8: An equilateral triangle is inscribed in the parabola y2=4ax, 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 a parabola with the equation.y2=4ax

One coordinate of the triangle is A(0,0).

Now, let the other two coordinates of the triangle be

B(x,4ax) and C(x,4ax)

Now, since the triangle is equilateral,

BC=AB=CA

24ax=(x0)2+(4ax0)2

x2=12ax

x=12a

The coordinates of the points of the equilateral triangle are,

(0,0),(12,4a×12a),(12,4a×12a)=(0,0),(12,43a)and(12,43a)

So, the side of the triangle is

24ax=2×43a=83a

Also read

Topics covered in Chapter 10 Conic Sections: Miscellaneous Exercise

1. Circle
A circle is the set of all points in a plane that are equidistant from a fixed point called the center.
(xh)2+(yk)2=r2
where (h,k) is the center and r is the radius.

2. Parabola
The collection of all points that are equally spaced between a given line and a fixed point (focus) is called a parabola.
Standard equation (horizontal axis) is given by
y2=4ax

3. Ellipse
An ellipse is the set of all points in a plane such that the sum of distances from two fixed points (called foci) is constant.
x2a2+y2b2=1

4. Hyperbola
The collection of all points in a plane where the distance difference between two fixed points (foci) is constant is called a hyperbola.
x2a2y2b2=1

5. Latus Rectum
A line segment that runs through the focus and is perpendicular to a conic's axis of symmetry is called the latus rectum. It helps in defining the width of a conic near the focus.

Also read

NCERT Solutions of Class 11 Subject Wise

Students can download the NCERT solutions from the link below and can make their studies more effective.

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Subject-Wise NCERT Exemplar Solutions

Take a step ahead in your preparations by following our NCERT exemplar solutions that are designed to give you a deeper understanding of the concepts.


Frequently Asked Questions (FAQs)

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

Equation of circle => (x1)2+(y2)2=52

x22x+1+y2+4y+4=25

x22x+y2+4y=20

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

Equation of the parabola =>  y2=4(2)x

y2=8x

3. Find the coordinates of the focus of the parabola y2=12x

 Given equation of the parabola  => y2=12x

Compare with y2=4ax   =>  a = 3

Coordinate of focus is (3,0).

4. Find the axis of the parabola y2=12x

The given equation involves y2, 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 y2=12x

Given equation of the parabola  => y2=12x

Compare with y2=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 y2=12x

Given equation of the parabola  => y2=12x

Compare with y2=4ax  =>  a = 3

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

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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