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NCERT Exemplar Class 11 Maths Solutions Chapter 11 Conic Sections

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NCERT Exemplar Class 11 Maths Solutions Chapter 11 Conic Sections

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:13 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 11 have been formulated by the professionals for the students to get detailed yet interesting insight into the concept of coordinate geometry in reference to curves. The chapter explains and covers a wider section of coordinate geometry dealing with the concept of curves through the previously learned concepts of geometry and algebra dealing with conic sections of circle, ellipse, parabola and hyperbola. NCERT exemplar Class 11 Maths solutions chapter 11 are provided by the professionals to provide efficient study material and help to students in order to assist them perfectly for any form of examination and application of the concept in different situations. The experts have used a very easy yet powerful approach for the NCERT exemplar Class 11 Maths solutions chapter 11 that make the learning process interesting yet knowledgeable.
Also, check - NCERT Class 11 Maths Solutions for Other Chapters

NCERT Exemplar Class 11 Maths Solutions Chapter 11: Exercise-1.3

Question:1

Find the equation of the circle which touches both axes in first quadrant and whose radius is a.
Answer:

Given that the circle has a radius aand touches both axis. So the centre is (a,a).
(x-a)2+(y-a)2=a2
x2+y2-2ax-2ay+a2=0

Question:3

If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.
Answer:

a-3
We have a circle through the point A(0,0), B(a,0) and C(0,b). Clearly triangle is right angled at vertex A. So, the centre of the circle is the mid-point of hypotenuse BC which is (a/2,b/2)

Question:4

Find the equation of the circle which touches x-axis and whose centre is (1, 2).

Answer:

It is given that circle with centre (1,2) touches x axis
Radius of the circle is r = 2
So equation of the required circle is
(x-1)2+(y-2)2=(2)2
x2-2x+1+y2-4y+4=4
x2+y2-2x-4y+1=0

Question:5

If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.

Answer:

Given lines are 6x-8y+8=0 and 6x-8y-7=0
These parallel lines are tangent to a circle
Diameter of circle = Distance between the lines
Diameter of circle=\left |\frac{8-\left ( -7 \right )}{\sqrt{36+64}} \right |=\frac{15}{10}=\frac{3}{2}
Radius of circle=\frac{3}{4}

Question:6

Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0and lies in the third quadrant.

Answer:

Since circle touches both the axes, its centre is C(-a, -a) and radius is a
Also, circle touches the line 3x-4y+8=0
Distance from centre C to this line is radius of the circle
Radius of circle, a=\left |\frac{-3a+4a+8}{\sqrt{9+16}} \right |=\left | \frac{a+8}{5} \right |
\left | \frac{a+8}{5} \right |=\pm a
a=2 or a=-4/3
a=2
Equation of required circle
(x+2)2+(y+2)2=22
x2+y2+4x+4y+4=0

Question:7

If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.

Answer:

Given equation of the circle is
x2+y2-4x-6y+11=0
2g=-4 and 2f=-6
So centre of circle is C(-g,-f) = C(2,3)
A (3,4) is one end of diameter . Let the other end of the diameter be B (x1, y1)
2=\frac{3+x_{1}}{2} and 3=\frac{4+y_{1}}{2}
x1=1 and y1=2

Question:8

Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18

Answer:

Given lines are
3x+y=14 and 2x+5y=18
Solving these equations we get point of intersection of the lines as A (4,2)
Radius= \sqrt{\left (4-1 \right )^{2}+\left (2-(-2 \right )^{2}}=\sqrt{9+16}=5
So, equation of the required circle is:
(x-1)2+(y+2)2=52
x2+y2-2x+4y-20=0

Question:9

If the line y=\sqrt{3}x+k touches the circle x2 + y2 = 16, then find the value of k.

Answer:

Given line is y=\sqrt{3}x+k and the circle is
x2+y2=16
Centre of the circle is (0,0) and radius is 4.
Since the line y=\sqrt{3}x+k touches the circle, perpendicular distance from (0,0) to line
is equal to the radius of the circle.
\left | \frac{0-0+k}{\sqrt{3+1}} \right |=4
\pm \frac{k}{2}=4
k=±8

Question:10

Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and has double of its area.

Answer:

Given equation of the circle is :
x2+y2-6x+12y+15=0
(x-3)2+(y+6)2=(30)
Hence, centre is (3, -6) and radius is \sqrt{30 }
Since, the required circle is concentric with above circle, centre of the required circle is (3, -6).
Let its radius be r
Area of circle=\Pi r^{2}=2\pi\left (\sqrt{30 } \right )^{2}
r2=60
r = \sqrt{60}
Equation of required circle (x-3)2+(y+6)2=60
x2+y2-6x+12y-15=0

Question:11

If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Answer:

Consider the equation of the ellipse as \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
It is given that length of latus rectum = Half of minor axis
\frac{2b^{2}}{a}=b
a=2b
b2=a2(1-e2)
b2=4b2(1-e2)
1-e2=1/4
e2=3/4
e=\frac{\sqrt{3}}{2}

Question:12

Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.

Answer:

Given equation of ellipse
9x2+25y2=225
\frac{x^{2}}{25}+\frac{y^{2}}{9}=1
a=5, b=3
b2=a2(1-e2)
9=25(1-e2)
e=\frac{4}{5}
\text { Foci } \equiv(\pm a e, 0) \equiv(\pm 5 \times(4 / 5), 0) \equiv(\pm 4,0)

Question:13

If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.

Answer:

Let the equation of the ellipse be \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
Distance between foci 2ae=10
It si given that eccentricity e=\frac{5}{8}
a=\frac{10}{2}*\frac{8}{5}=8
b=\sqrt{a^{2}\left ( 1-e^{2} \right )}=\sqrt{39}
Length of latus rectum of ellipse=\frac{2b^{2}}{a}=2*\frac{39}{8}=\frac{39}{4}

Question:14

Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and the centre is (0, 0).

Answer:

Let the equation of the ellipse be \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
e=\frac{2}{3}
Given that latus rectum \frac{2b^{2}}{a}=5
b^{2}=\frac{5a}{2}=a^{2}\left ( 1-e^{2} \right )
\frac{5}{2}=a\left ( 1-\left ( \frac{2}{3} \right )^{2} \right )
a=\frac{9}{2}
b^{2}=\frac{45}{4}
The required equation of the ellipse is\frac{4x^{2}}{81}+\frac{4y^{2}}{45}=1

Question:15

Find the distance between the directrices of the ellipse \frac{x^{2}}{36}+\frac{y^{2}}{20}=1

Answer:

The equation of the ellipse is \frac{x^{2}}{36}+\frac{y^{2}}{20}=1
a=6 andb=2\sqrt{5}
We know that b^{2}=a^{2}\left ( 1-e^{2} \right )
\frac{20}{36}=1-e^{2}
e^{2}=\frac{4}{9}
x=\pm \frac{a}{e}
Distance between directrix=\frac{2a}{e}=\frac{2*6}{\frac{2}{3}}=18

Question:16

Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4.

Answer:

Given parabola is y2=8x
On comparing this parabola to the y2=4ax
we get a=2
Focal distance=Distance of any point on parabola from the focus SP=\sqrt{\left ( x_{1}-2 \right )^{2}+\left ( y_{1}-0\right )^{2}}
\sqrt{ x_{1}^{2}+ y_{1}^{2}-4x_{1}+4}+\sqrt{ x_{1}^{2}+ 8x_{1}-4x_{1}+4}=\left | x_{1}+2 \right |
|x1+2|=4
x1=2,-6
But x≠ -6
So for x=2
y_{1}^{2}=8*2=16
y= ±4 So the points are (2,4) and 2,-4

Question:18

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Answer:

sdfghjjhdfghjk
Given that the vertex of the parabola is A(0,4) and its focus is S(0,2)
So, directrix of the parabola is y=6
Now by definition of the parabola for any point P (x,y) on the parabola SP=PM
\sqrt{\left ( x-0 \right )^{2}+\left ( y-2 \right )^{2}}=\left | \frac{0+y-6}{\sqrt(0+1)} \right |
x2+y2-4y+4=y2-12y+36
x2+8y=32

Question:19

If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m.

Answer:

Given that line y=mx+1 is tangent to the parabola y2=4x
Solving line with parabola we have
(mx+1)2=4x
m2x2+2mx+1=4x
m2x2+x(2m-4)+1=0
Since the line touches the parabola, above equation must have equal roots.
Discriminant D=0
(2m-4)2-4m2=0
4m2-16m+16-4m2=0
16m=16
m=1

Question:20

If the distance between the foci of a hyperbola is 16 and its eccentricity is \sqrt{2}, then obtain the equation of the hyperbola
Answer:

Let the equation of the hyperbola be \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
Distance between foci=2ae=16
e=\sqrt{2}
a=4\sqrt{2}
We know that b=\sqrt{a^{2}\left ( 1-e^{2} \right )}=\sqrt{32}
\frac{x^{2}}{32}-\frac{y^{2}}{32}=1
x2-y2=32

Question:21

Find the eccentricity of the hyperbola 9y2 – 4x2 = 36

Answer:

We have the hyperbola 9y2-4x2=36
\frac{x^{2}}{9}-\frac{y^{2}}{4}=-1
a2=9
b2=4
a2=b2(e2-1)
e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}

Question:22

Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).

Answer:

Let the equation of the hyperbola be \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
ae=2
a*\frac{3}{2}=2
a=\frac{4}{3}
b2=a2(e2-1)
b^{2}=\frac{16}{9}*\left ( \frac{9}{4}-1 \right )=\frac{20}{9}
\frac{x^{2}}{\frac{16}{9}}-\frac{y^{2}}{\frac{20}{9}}=1
\frac{x^{2}}{16}-\frac{y^{2}}{20}=\frac{1}{9}

Question:23

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

Answer:

Given that the lines 2x-3y-5=0 and 3x-4y-7=0 are diameters of the circle.
2x-3y-5=0.....(i)
3x-4y-7=0....(ii)
3 (Equation (i)) - 2 (Equation (ii) )
6x-9y-15-6x+8y+14=0.
y = -1
Put value of y in equation (i)
x = 1
Solving these lines we will get the intersection as (1,-1) which is centre of the circle.
Also, given that area of the circle is 154 sq units
πr2=154
r2=154*7/22=49
r=7
(x-1)2+(y+1)2=49
x2+y2-2x+2y-47=0

Question:24

Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.

Answer:

Let the centre of the circle be C(h,k)
Given that the centre lies on the line y-4x+3=0
y-4x+3=0
k-4h+3=0
k=4h-3
AC2=BC2
(h-2)2+(4h-3-3)2=(h-4)2+(4h-3-5)2
h2-4h+4+16h2-48h+36=h2-8h+16+16h2-64h+64
20h=40
h=2
k=4h-3=5
radius=\sqrt{\left ( 2-2 \right )^{2}+\left ( 3-5 \right )^{2}}=2
Therefore equation of the circle is : (x-2)2+(y-5)2=4
x2+y2-4x-10y+25=0

Question:25

Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0.

Answer:

sdfghjkl
Given centre of the circle O(3, -1)
Chord of the circle is AB
Given that equation of AB is 2x-5y+18=0
Perpendicular distance from O to AB is OP= \left | \frac{2(3)-5(-1)+18}{\sqrt{4+25}} \right |=\frac{29}{\sqrt{29}}=\sqrt{29}
OB2=OP2+PB2
OB2=29+9=38
OB=\sqrt{38}
Equation of the circle is (x-3)2+(y+1)2=38
x2+y2-6x+2y=28

Question:26

Find the equation of a circle of radius 5 which is touching another circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5).

Answer:

Given circle is x2+y2-2x-4y=20
(x-1)2+(y-2)2=52
Centre of this circle is C1(1,2).
Now, the required circle of radius'5'touches the above circle at P(5,5).
Let the centre of the required circle be C2(h,k).
Since the radius of the given circle and the required circle is same, point P is mid point of C1C2.
5=\frac{1+h}{2} and 5=\frac{2+k}{2}
h=9 and k=8
So the equation of the required circle is (x-9)2+(y-8)2=25
x2+y2-18x-16y+120=0

Question:27

Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.

Answer:

Given that circle passes through the point A(7,3) and its radius is 3
Therefore centre of the circle is C(h,h-1)
Now, radius of the circle is AC=3
(h-7)2+(h-1-3)2=32
2h2-22h+56=0
h2-11h+28=0
(h-4)(h-7)=0
h=4 or 7
If the centre of the circle is C(4,3)⇒(x-4)2+(y-3)2=9⇒x2+y2-8x-6y+16=0
If the centre is C(7,6)⇒(x-7)2+(y-6)2=9⇒x2+y2-14x-12y+76=0

Question:28

Find the equation of each of the following parabolas
(a) Directrix x = 0, focus at (6, 0)
(b) Vertex at (0, 4), focus at (0, 2)
(c) Focus at (–1, –2), directrix x – 2y + 3 = 0

Answer:

We know that the distance of any point on the parabola from its focus and its directrix is same.
i) Given that directrix x=0 and focus (6,0)
So, for any point P(x,y) on the parabola
Distance of P from directrix=Distance of P from focus x2=(x-6)2+y2
y2-12x+36=0
ii) Given that vertex=(0,4) and focus (0,2)
Now distance between the vertex and directrix is same as the distance between the vertex and focus.
Distance of P from directrix=Distance of P from focus
\left | y-6 \right |=\sqrt{\left ( x-0 \right )^{2}+\left ( y-2 \right )^{2}}
y2-12y+36=x2+y2-4y+4
x2=32-8y
iii) Given that focus is at (-1,-2)
and directrix x-2y+3=0
\sqrt{\left ( x+1 \right )^{2}+\left ( y+2 \right )^{2}}=\left | \frac{x-2y+3}{\sqrt{1+4}} \right |
x2+2x+1+y2+4y+4=1/5[x2+4y2+9+6x-4xy-12y]
4x2+4xy+y2+4x+32y+16=0

Question:29

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Answer:

Let the coordinates of the variable point be (x,y)
Then according to the question \sqrt{\left ( x-3 \right )^{2}+y^{2}}=12-\sqrt{\left ( x-9 \right )^{2}+y^{2}}
x2-6x+9+y2=144+x^{2}-18x+81+y^{2}-24\sqrt{\left (x-9 \right )^{2}+y^{2}}
\left ( x-18 \right )=-2\sqrt{\left (x-9 \right )^{2}+y^{2}}
Again squaring both sides, we get
x2-36x+324=4(x2-18x+81+y2)
3x2+4y2-36x=0 which is an ellipse.

Question:30

Find the equation of the set of all points whose distance from (0, 4) are 2/3 of their distance from the line y = 9.

Answer:

Let the point be P (x,y)
According to the question
\sqrt{x^{2}+\left ( y-4 \right )^{2}}=\frac{2}{3}\left | \frac{y-9}{1} \right |
9(x2+y2-8y+16)=4(y2-18y+81)
9x2+5y2=180

Question:31

Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represent a hyperbola.

Answer:

Let the point be P (x,y)
According to the question
Distance of P from (-4,0) -Distance of P from (4,0)=2
\sqrt{\left ( x+4 \right )^{2}+y^{2}}=2+\sqrt{\left ( x-4 \right )^{2}+y^{2}}
Squaring both the sides
x^{2}+8x+16+y^{2}=4+x^{2}-8x+16+y^{2}+4\sqrt{\left ( x-4 \right )^{2}+y^{2}}
4x-1=\sqrt{\left ( x-4 \right )^{2}+y^{2}}
16x2-8x+1=x2+16-8x+y2
15x2-y2=15
Which is an equation of a hyperbola.

Question:32

Find the equation of the hyperbola with

(i) Vertices (± 5, 0), foci (± 7, 0)
(ii) Vertices (0, ± 7), e = \frac{4}{3}
(iii) Foci \left ( 0,\pm \sqrt{10} \right ), passing through (2, 3)

Answer:

a) Given that a=5 and ae=7
e=7/5
b2=a2(e2-1)=25(49/25-1)=24
So, the equation of hyperbola is
\frac{x^{2}}{25}-\frac{y^{2}}{24}=1
b) b=7,e=4/3
a^{2}=b^{2}\left ( e^{2} - 1\right )=\frac{343}{9}
\frac{x^{2}}{\frac{343}{9}}-\frac{y^{2}}{49}=-1
9x^{2}-7y^{2}+343=0
c) Given that foci=(0,±10 ) be= \sqrt{10}
a2=b2(e2-1)
a2=b2e2-b2
a2=10-b2
\frac{x^{2}}{10-b^{2}}-\frac{y^{2}}{b^{2}}=-1 Since, hyperbola passes through the point (2,3)
4b2-9(10-b2)=-b2(10-b2)
b4-23b2+90=0
b4-18b2-5b2+90=0
(b2-18)(b2-5)=0
b2=5
a2+b2=10⇒a2=5
\frac{x^{2}}{5}-\frac{y^{2}}{5}=-1
y2-x2=5

Question:33

The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0.

Answer:

False
Given equation of the circle is x2+y2+6x+2y=0
Centre=(-3,-1)
Clearly, it does not lie on the line x+3y=0 as-3+3(-1)=-6
So this line is not the diameter of the circle

Question:35

If the line lx + my = 1 is a tangent to the circle x2 + y2 = a2, then the point (l, m) lies on a circle.

Answer:

True
Given circle is x2+y2=a2
The given line lx+my-1=0 is tangent to the circle
Distance of (0,0) from the line lx+my-1=0 is equal to the radius a
\left | \frac{0+0+1}{\sqrt{l^{2+m^{2}}}} \right |=a
l^2+m^{2}=\frac{1}{a^{2}}
The locus of (l,m) is x^{2}+y^{2}=\frac{1}{a^{2}}. which is an equation of a circle.

Question:36
The point (1, 2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0.

Answer:

False
Given circle is x2+y2-2x+6y+1=0 or (x-1)2+(y+3)2=32
Centre is C(1,-3) and radius is 3.
Distance of point P (1,2) from centre is 5.
Thus CP>radius
So, point P lies outside the circle.

Question:37

The line lx + my + n = 0 will touch the parabola y2 = 4ax if ln = am2.

Answer:

True
Give line lx+my+n=0
and parabola y2=4ax
Solving line and parabola for their point of intersection we get
l\left ( \frac{y^{2}}{4a} \right )+my+n=0
Since line touches the parabola, above equation must have equal roots
D=0
m^{2}-4*\left ( \frac{1}{4a} \right )*n=0
am2=nl

Question:38

If P is a point on the ellipse \frac{x^{2}}{16}+\frac{x^{2}}{25}=1 whose foci are S and S′, then PS + PS′ = 8.
Answer:

False
We have definition of the ellipse is \frac{x^{2}}{16}+\frac{y^2}{25}=1
From the definition of the ellipse, we know that sum of the distances of any point P
on the ellipse from the two foci is equal to the length of the major axis.
Here major axis=2b=2*5=10
S and S'are foci, then SP+S'P=10

Question:39

The line 2x + 3y = 12 touches the ellipse\frac{x^{2}}{9}+\frac{y^{2}}{4}=2 at the point (3, 2).

Answer:

True
Given line is 2x+3y=12
and the ellipse 4x2+9y2=72
Solving line and ellipse we get (12-3y)2+9y2=72
(4-y)2+y2=8
2y2-8y+8=0
y2-4y+4=0
(y-2)2=0
y=2;x=3

Question:40

The locus of the point of intersection of lines \sqrt{3}x-y-4\sqrt{3}k=0 and \sqrt{3}kx+ky-4\sqrt{3}=0 for different value of k is a hyperbola whose eccentricity is 2.

Answer:

True
Given equation of lines are \sqrt{3}x-y-4\sqrt{3}k=0
\sqrt{3}kx+ky-4\sqrt{3}=0
k=\frac{\sqrt{3}x-y}{4\sqrt{3}} andk=\frac{4\sqrt{3}}{\sqrt{3}x+y}
\frac{\sqrt{3}x-y}{4\sqrt{3}} =\frac{4\sqrt{3}}{\sqrt{3}x+y}
3x2-y2=48
\frac{x^{2}}{16}-\frac{y^{2}}{48}=1 which is the equation of hyperbola
a2=16 and b2=48
e2=1+48/16=4
e=2

Question:41

The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is ______.

Answer:

The perpendicular distance from centre (3,-4) to the given line is,r=\left | \frac{5(3)+12(-4)-12}{\sqrt{25-144}} \right |
=\frac{45}{13} which is radius of the circle
Required equation of the circle is \left ( x-3 \right )^{2}+\left ( y+4 \right )^{2}=\left (\frac{45}{13} \right )^{2}

Question:42

The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is __________ .

Answer:

Given equation of line are y=x+2
3y=4x
2y=3x
Solving these lines, we get points of intersection A(6,8), B (4,6) and C(0,0).
Let the equation of circle circumscribing the given triangle be
x2+y2+2gx+2fy+c=0
36+64+12g+16f+c=0⇒12g+16f+c=-100
16+36+8g+12f+c=0⇒8g+12f+c=-52
0+0+0+0+c=0⇒c=0
3g+4f=-25
2g+3f=-13
On solving we get g=-23 and f=11
The equation of circle is x2+y2-46x+22y=0

Question:43

An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are ____________.

Answer:

Let equation of ellipse be \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
a=3 and b=2
b2=a2(1-e2)
e2=1-4/9=5/9
e=\frac{\sqrt{5}}{3}
From the definition of the ellipse for any point P on the ellipse we have
SP+S'P=2a
Length of endless string=SP+S'P+SS'=2a+2ae=2(3)+2(3)*\frac{\sqrt{5}}{3}=6+2\sqrt{5}

Question:44

The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is_________

Answer:

be=1
Length of minor axis 2a=1
a=1/2
a2=b2(1-e2)
1/4=b2-b2e2
b2=1/4+1=5/4
\frac{x^{2}}{\frac{1}{4}}+\frac{y^{2}}{\frac{5}{4}}=1
So, the equation of ellipse is 4x^+\frac{4y^{2}}{5}=1

Question:45

The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is ____________ .

Answer:

Let any point on parabola be P(x,y)
Length of perpendicular from S on the directrix=SP
\frac{\left ( x-2y+3 \right )^{2}}{5}=\left ( x+1 \right )^{2}+\left ( y+2 \right )^{2}
4x2+y2+4x+32y+16=0

Question:47

The area of the circle centred at (1, 2) and passing through (4, 6) is
A. 5π
B. 10π
C. 25π
D. none of these

Answer:

Centre of the circle is C(1,2)
Radius=\sqrt{\left ( 4-1 \right )^{2}+\left ( 6-2 \right )^{2}}=5
Area of circle= π(5)2=25π square units

Question:48

Equation of a circle which passes through (3, 6) and touches the axes is
A. x2 + y2 + 6x + 6y + 3 = 0
B. x2 + y2 – 6x – 6y – 9 = 0
C. x2 + y2 – 6x – 6y + 9 = 0
D. none of these

Answer:

Given that the circle touches both axes
Therefore, equation of the circle is (x-a)2+(y-a)2=a2
circle passes through (3,6)
(3-a)2+(6-a)2=a2
a2-18a+45=0
(a-3)(a-15)=0
a = 3 or a =15
For a=3 equation of circle
(x-3)2+(y-3)2=9
x2+y2-6x-6y+9=0

Question:49

Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
A. x2 + y2 + 13y = 0
B. 3x2 + 3y2 + 13x + 3 = 0
C. 6x2 + 6y2 – 13x = 0
D. x2 + y2 +13x +3 = 0

Answer:

Centre of the circle lies on the y axis
So, let the centre be C(0,k)
Circle passes through O(0,0)and A(2,3)
OC2=AC2
k2=(2-0)2+(3-k)2
k=13/6
(x-0)2+(y-13/6)2=(13/6)2
3x2+3y2-13y=0
Answer- None

Question:50

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
A. x2 + y2 = 9a2
B. x2 + y2 = 16a2
C.x2 + y2 = 4a2
D. x2 + y2 = a2

Answer:

Option (C)
median length=3a
Radius of circle=2/3*median length=2a
Equation of circle
x2+y2=4a2

Question:51

If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is
A. x2 = –12y
B. x2 = 12y
C. y2 = –12x
D. y2 = 12x

Answer:

a Given that, focus of parabola is at S(0,-3)and equation of directrix is y=3
For any point P(x,y)on the parabola, we have
SP=PM
\sqrt{(x-0)^{2}+(y+3)^{2}}=\left | y-3 \right |
x2+y2+6y+9=y2-6y+9
x2=-12y

Question:52

If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is
A. 2/3
B. 4/3
C. 1/3
D. 4

Answer:

Option (b) is correct.
Parabola y2=4ax, passes through the point (3,2)
4=4a(3)
a=1/3
Length of latus rectum 4a=4/3

Question:53

If the vertex of the parabola is the point (–3, 0) and the directrix is the line x + 5 = 0, then its equation is
A. y2 = 8 (x + 3)
B. x2 = 8 (y + 3)
C. y2 = – 8 (x + 3)
D. y2 = 8 ( x + 5)

Answer:

Given: Vertex =(–3, 0) and the directrix x = -5(M)
So focus S (-1,0)
For any point of parabola P (x,y) we have
SP=PM
\sqrt{(x+1)^{2}+y^{2}}=\left | x+5 \right |
x2+y2+2x+1=x2+10x+25
y2=8x+24

Question:54

The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity 1/2 is
A. 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0
B. 7x2 + 2xy + 7y2 + 7 = 0
C. 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0
D. none

Answer:

Option (a) is correct.
Given that focus of the ellipse is S(1,-1)and the equation of directrix is x-y-3=0
Also, e=1/2
From definition of ellipse, for any point P (x,y) on the ellipse, we have SP=ePM, where
M is foot of the perpendicular from point P to the directrix.
\sqrt{(x-1)^{2}+(y+1)^{2}}=\frac{1}{2}\frac{\left | x-y-3 \right |}{\sqrt{2}}
7x2+7y2+2xy-10x+10y+7=0

Question:55

The length of the latus rectum of the ellipse 3x2 + y2 = 12 is
A. 4
B. 3
C. 8
D. 12
Answer:

Option (d) is correct.
Given ellipse is 3x2+y2=12
\frac{x^{2}}{4}+\frac{y^{2}}{12}=1
a2=4
a=2
b2=12
b=2\sqrt{3}
length of latus rectum=\frac{2a^{2}}{b}=\frac{4}{\sqrt{3}}

Question:56

If e is the eccentricity of the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\left ( a<b \right ), then
A. b2 = a2 (1 – e2)
B. a2 = b2 (1 – e2)
C. a2 = b2 (e2 – 1)
D. b2 = a2 (e2 – 1)

Answer:

Option (b) is correct.
Given that \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
We know that a2=b2(1-e2)

Question:57

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is
A. 4/3

B. \frac{4}{\sqrt{3}}
C. \frac{2}{\sqrt{3}}
D. none of these

Answer:

Let the equation of the hyperbola be x2/a2-y2/b2=1
length of latus rectum=8
2b2/a=8
b2=4a
Conjugate axis=half of the distance between the foci
2b=ae
b2=a2(e2-1)
\frac{a^{2}e^{2}}{4}=a2(e2-1)
e2=4/3
e=\frac{2}{\sqrt{3}}

Question:58

The distance between the foci of a hyperbola is 16 and its eccentricity is \sqrt{2}. Its equation is
A. x2 – y2 = 32
B. \frac{x^{2}}{4}-\frac{y^{2}}{9}=1
C. 2x – 3y2 = 7
D. none of these

Answer:

Option (A) is correct.
Let the equation of the hyperbola be \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
Given that Foci=(±ae,0)
Distance between foci=2ae=16
2*a*\sqrt{2}=16
a=4\sqrt{2}
b^{2}=a^{2}(e^{2}-1)=(4\sqrt{2})^{2}((\sqrt{2})^{2}-1)
=32(2-1)=32
Hence, equation is \frac{x^{2}}{32}-\frac{y^{2}}{32}=1

Question:59

Equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0) is
A. \frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}
B. \frac{x^{2}}{9}-\frac{y^{2}}{9}=\frac{4}{9}
C. \frac{x^{2}}{4}-\frac{y^{2}}{9}=1
D. None of these

Answer:

Option (A) is correct.
Let the equation of the hyperbola be \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
e=3/2
Foci=(±ae),0=(±2,0)
So, after comparing the equations, ae=2
a*3/2=2
a=4/3
b2=a2(e2-1)
b2=(4/3)2((3/2)2-1)
=(16/9)(9/4-1)
=16/9*5/4=20/9
Equation of hyperbola is \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
=\frac{x^{2}}{\left (\frac{4}{3} \right )^{2}}-\frac{y^{2}}{\frac{20}{9}}=1
\frac{x^{2}}{\frac{16}{9}}-\frac{y^{2}}{\frac{20}{9}}=1
Hence, \frac{x^{2}}{16}-\frac{y^{2}}{{20}}=\frac{1}{9} is the required equation.

More About NCERT Exemplar Class 11 Maths Solutions Chapter 11

The students aiming at achieving the best possible solutions for NCERT questions can access the study material through NCERT Exemplar Class 11 Maths solutions chapter 11 PDF download that is drafted by professionals for expert guidance.

NCERT Exemplar solutions for Class 11 Maths chapter 11 help students for getting an easy yet effective approach to NCERT problems that are important for their exams and carefully drafted by studying the CBSE guidelines for problem-solving.

Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 11

  • 11.1 Introduction
  • 11.2 Sections of a cone
  • 11.2.1 Circle, ellipse, parabola and hyperbola
  • 11.2.2 Degenerated Conic section
  • 11.3 Circle
  • 11.4 Parabola
  • 11.4.1 Standard equations of parabola
  • 11.4.2 Latus Rectum
  • 11.5 Ellipse
  • 11.5.1 Relationship between semi-major axis, semi-minor axis and the distance of the focus from the center of the ellipse
  • 11.5.2 Special cases of an ellipse
  • 11.5.3 Eccentricity
  • 11.5.4 Standard equation of an ellipse
  • 11.5.5 Latus Rectum
  • 11.6 Hyperbola
  • 11.6.1 Eccentricity
  • 11.6.2 Standard equation of a hyperbola

What will the students learn in NCERT Exemplar Class 11 Maths Solutions Chapter 11?

With Class 11 Maths NCERT exemplar solutions chapter 11, the students will be introduced to the study of a conic section in a two-dimensional plane through the use of algebra and geometry that has a number of uses in various other fields of science.

  • The students interested in pursuing any field or branch of science in future will be guided a lot through NCERT Exemplar Class 11 Maths solutions chapter 11 as it has wide application in different fields including physics for understanding and examination of different laws and every other analysis and derivation along with the subject.
  • It also represents great ways to display the relationship between analytical and synthetic thinking in mathematics through the study to coordinate planes.
  • The chapter also aids in the introduction of matrices and determinants and their application for further study of curves and multivariable calculus.
  • The students will also learn about the geometry of different conic sections, including circle, ellipse, parabola and hyperbola along with the knowledge of equations of varied conic sections.

NCERT Exemplar Class 11 Maths Solutions Chapter-Wise

Important topics to cover from NCERT Exemplar Class 11 Maths Solutions Chapter 11

Some of the important topics for students to review from this chapter are:

· The students will learn about the representation of curved figures such as circle, ellipse, hyperbola and parabola in the coordinate plane.

· NCERT Exemplar Class 11 Maths solutions chapter 11 covers standard equations of circle and parabola.

· The students must cover standard equations of parabola and hyperbola in addition to their eccentricity and latus rectrum.

· The students must thoroughly practice the solutions and examples of NCERT for examination.

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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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Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available
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