Have you ever thought about how a satellite dish is designed or how one decides the orbit of planets? Conic sections hold the answer! NCERT Exemplar Class 11 Maths Chapter 11 Conic Sections is a fundamental and significant chapter. The student learns about several shapes produced when a slanted cut is made through a cone in this chapter; these shapes include circles, parabolas, ellipses, and hyperbolas. It also covers knowledge about their usual forms, most essential qualities, and daily uses in disciplines including physics, astronomy, and architecture. Students will also have knowledge about eccentricity, latus rectum, and focal distances, all of which are necessary for their mathematical knowledge and in future chapters.
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This article provides detailed and easy-to-understand Solutions for all NCERT Exemplar questions of Class 11 Maths Chapter 11, curated by Careers360 experts. You can check the complete Class 11 Maths syllabus through the NCERT Syllabus Class 11 Maths and explore Solutions for other chapters in the NCERT Class 11 Maths Solutions. Check this NCERT article for complete syllabus coverage along with NCERT Books, Solutions, Syllabus, and Exemplar Problems with Solutions.
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| Class 11 Maths Chapter 11 Exemplar Solutions Exercise: 11.3 Page number: 202-207 Total questions: 59 |
Question 1
Find the equation of the circle which touches both axes in the first quadrant and whose radius is a.
Answer:
Given that the circle has a radius and touches both axes, the centre is (a, a).
$\begin{aligned} & (x-a)^2+(y-a)^2=a^2 \\ & x^2+y^2-2 a x-2 a y+a^2=0\end{aligned}$
Question 2
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:
We have a circle through the points A(0,0), B(a,0) and C(0,b). The triangle is right-angled at vertex A. So, the centre of the circle is the mid-point of hypotenuse B, C, which is (a/2,b/2)
Question 4
Find the equation of the circle which touches the x-axis and whose centre is (1, 2).
Answer:
It is given that a circle with a centre (1,2) touches the x-axis
The radius of the circle is r = 2
The equation of the required circle is:
$\begin{aligned} & (x-1)^2+(y-2)^2=(2)^2 \\ & x^2-2 x+1+y^2-4 y+4=4 \\ & x^2+y^2-2 x-4 y+1=0\end{aligned}$
Question 5
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
Answer:
Since the circle touches both axes, its centre is C(-a, a) and its radius is a
Also, the circle touches the line 3x-4y+8=0
The distance from centre C to this line is the 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 a required circle
(x+2)2+(y+2)2=22
x2+y2+4x+4y+4=0
Question 7
Answer:
Given the 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 the 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
Answer:
Given lines are
3x+y=14 and 2x+5y=18
Solving these equations, we get the 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, the 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=16The The The
The Centre of the circle is (0,0) and the radius is 4.
Since the line $y=\sqrt{3}x+k$ touches the circle, the perpendicular distance from (0,0) to the 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
Answer:
The 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 the above circle, the 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 the 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 the length of the latus rectum = Half of the minor axis
$\frac{2b^{2}}{a}=b$
a=2b
b2=a2(1-e2)
b2=4b2(1-e2)
1-e2=$\frac 14$
e2=$\frac 34$
$e=\frac{\sqrt{3}}{2}$
Question 12
Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.
Answer:
Given the equation of an 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
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 is 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
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 the 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
Answer:
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
Answer:
Given that the vertex of the parabola is A(0,4) and its focus is S(0,2)
So, the directrix of the parabola is y=6
Now, by definition of the parabola, for any point P (x,y) on the parabola, a 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 the line with a parabola, we have
(mx+1)2=4x
m2x2+2mx+1=4x
m2x2+x(2m-4)+1=0
Since the line touches the parabola, the 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
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
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 the value of y in equation (i)
x = 1
Solving these lines, we will get the intersection as (1,-1), which is the centre of the circle.
Also, given that the 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
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, the equation of the circle is : (x-2)2+(y-5)2=4
x2+y2-4x-10y+25=0
Question 25
Answer:
Given centre of the circle O(3, -1)
The chord of the circle is AB
Given that the 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
Answer:
The 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 are the same, point P is the midpoint 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
Answer:
Given that the circle passes through the point A(7,3) and its radius is 3
Therefore, the centre of the circle is C(h,h-1)
Now, the 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
Answer:
We know that the distance of any point on the parabola from its focus and its directrix is the same.
i) Given that the directrix x=0 and the 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 the distance between the vertex and the directrix is the same as the distance between the vertex and the 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 the 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
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
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
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 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
Answer:
a) Given that a=5 and ae = 7
e=$\frac 75$
b2=a2(e2-1)=25(49/25-1)=24
So, the equation of the 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 a 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 athediameter of the circle x2 + y2 + 6x + 2y = 0.
Answer:
False
Given equation of the circle is x2+y2+6x+2y=0
Centre=(-3,-1)
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
False
$
\begin{aligned}
& \text { Centre } \equiv C(7,5) \\
& \text { Radius }=\sqrt{49+25+151}=\sqrt{225}=15
\end{aligned}
$
Distance between the point $P(2,-7)$ and centre
$
=\sqrt{(2-7)^2+(-7-5)^2}=\sqrt{169}=13
$
The shortest distance of point P from the circle $=|13-15|=2$
Question 35
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 the radius is 3.
The distance of point P (1,2) from the 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 the line and parabola for their point of intersection, we get
$l\left ( \frac{y^{2}}{4a} \right )+my+n=0$
Since the line touches the parabola, the above equation must have equal roots
D=0
$m^{2}-4×\left ( \frac{1}{4a} \right )×n=0$
am2=nl
Question 38
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 the 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 the line and the 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
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
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
Answer:
Given equation of the line is 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 the 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 a circle is x2+y2-46x+22y=0
Question 43
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)
$\frac 14$=b2-b2e2
b2=$\frac 14$+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
Answer:
Let any point on the parabola be P(x,y)
Length of the perpendicular from S to the directrix
$\frac{\left ( x-2y+3 \right )^{2}}{5}=\left ( x+1 \right )^{2}+\left ( y+2 \right )^{2}$
4x2+y2+4x+32y+16=0
Question 46
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
The 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, the 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 an =3 equation of a circle
(x-3)2+(y-3)2=9
x2+y2-6x-6y+9=0
Question 49
The equation of the circle with the 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
Solution:
The 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 a 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 the 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 the 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 = 0D. 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 the root 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 whose 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
The 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.
Students can easily find all NCERT Class 11 Maths Exemplar Solutions in one place on Careers360. Simply tap the links below to get started.
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Solving the NCERT Exemplar Class 11 Maths Chapter 11 helps students develop a strong understanding of conic sections and their properties. These problems improve conceptual clarity and problem-solving skills, which are essential for exams and higher-level mathematics. Regular practice also boosts confidence in tackling coordinate geometry questions.
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Option 2)
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Option 4)
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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
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Option 2)
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| Option 3)
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Option 4)
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
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Option 2)
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Option 4)
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In the reaction,
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Option 2)
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Option 4)
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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
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0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
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Option 4)
2.5 × 10-2 |
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