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.
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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:2

Show that the point (x, y) given by $x=\frac{2at}{1+t^{2}}$ and $x=\frac{a\left ( 1-t^{2} \right )}{1+t^{2}}$ lies on a circle for all real values of t such that –1 ≤ t ≤1 where a is any given real numbers.

Answer:

We have variable point as $x=\frac{2at}{1+t^{2}}$ and $y=\frac{a\left ( 1-t^{2} \right )}{1+t^{2}}$
$\\x^{2}+y^{2}=\frac{4a^{2}t^{2}}{\left ( 1+t^{2} \right )^{2}}+\frac{a^{2}\left ( 1+t^{4}-2t^{2} \right )}{\left ( 1+t^{2} \right )^{2}} \\\\=\frac{a^{2}+a^{2}t^{4}+2a^{2}t^{2}}{\left ( 1+t^{2} \right )^2}\\\\=\frac{a^{2}\left ( 1+t^{2} \right )^{2}}{\left ( 1+t^{2} \right )^2}\\\\= a^{2}$

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 and $b=2\sqrt{5}$
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:17

Find the length of the line-segment joining the vertex of the parabola y² = 4ax and a point on the parabola where the line-segment makes an angle $\theta$ to the x-axis.

Answer:

a17

Let the point on the parabola be P (x₁,y₁)
Slope of OP= $\tan \theta =\frac{y_{1}}{x_{1}}$
$y_{1}^{2}=4ax_{1}$
$x_{1}=\frac{4a}{\tan ^{2}\theta }$
$OP=\frac{4a \sec \theta }{\tan ^{2}\theta }= \frac{4a \cos \theta }{\sin^{2}\theta }$

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:34

The shortest distance from the point (2, –7) to the circle x²+ y²– 14x – 10y – 151 = 0 is equal to 5.

Answer:

False
$\text { Centre } \equiv C(7,5) \\$
$$ Radius $=\sqrt{49+25+151}=\sqrt{225}=15$
$\\$Distance between the point $P(2,-7)$ and centre $$ =\sqrt{(2-7)^{2}+(-7-5)^{2}}$ = $\sqrt{169}=13$
$\text {The shortest distance of point } \mathrm{P} \text{ from the circle} =|13-15|=2$

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 $\sqrt{3}x-y-4\sqrt{3}k=0$
$\sqrt{3}kx+ky-4\sqrt{3}=0$
$k=\frac{\sqrt{3}x-y}{4\sqrt{3}}$ and $k=\frac{4\sqrt{3}}{\sqrt{3}x+y}$
$\frac{\sqrt{3}x-y}{4\sqrt{3}}$ $=\frac{4\sqrt{3}}{\sqrt{3}x+y}$
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:46

The equation of the hyperbola with vertices at (0, ± 6) and eccentricity 5/3 is ________ and its foci are __________ .

Answer:

Let the equation of parabola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1$
b=6
e=5/3
a²=b²(e²-1)
a²=64
$\frac{x^{2}}{64}-\frac{y^{2}}{36}=-1$
$\text { foci }=(0, \pm b e) \equiv\left(0, \pm \frac{5}{3} \times 6\right)=(0,\pm 10)$

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

Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 3 Trigonometric Functions

Chapter 4 Principle of Mathematical Induction

Chapter 5 Complex Numbers and Quadratic Equations

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 9 Sequences and Series

Chapter 10 Straight lines

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 15 Statistics

Chapter 16 Probability

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.

Check Chapter-Wise NCERT Solutions of Book

Chapter-1	Sets
Chapter-2	Relations and Functions
Chapter-3	Trigonometric Functions
Chapter-4	Principle of Mathematical Induction
Chapter-5	Complex Numbers and Quadratic equations
Chapter-6	Linear Inequalities
Chapter-7	Permutation and Combinations
Chapter-8	Binomial Theorem
Chapter-9	Sequences and Series
Chapter-10	Straight Lines
Chapter-11	Conic Section
Chapter-12	Introduction to Three Dimensional Geometry
Chapter-13	Limits and Derivatives
Chapter-14	Mathematical Reasoning
Chapter-15	Statistics
Chapter-16	Probability

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NCERT Exemplar Class 11 Maths chapter 11 solutions provided on our website are created by a trained professional of mathematics that provides a better understanding and easier methods through research of various problem-solving methods suited to CBSE.

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Yes, the NCERT exemplar solutions for Class 11 Maths chapter 11 developed by the experts are very reliable from the perspective of exams and are drafted in a way to satisfy the methods and guidelines of CBSE and NCERT.

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NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
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Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
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NCERT Exemplar Class 11 Maths Solutions Chapter 11 Conic Sections

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

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