CBSE Solutions for Class 11 Maths

Select CBSE Solutions for class 10 Subject & Chapters Wise :

If the equation of a circle is 2λx2 + (4λ − 6)y2 − 8x + 12y − 2 = 0, then the coordinates of centre are _________

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Answer :

(2/3, −1)

The equation x2 + y2 + 2x − 4y + 5 = 0 represents _____________

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Answer :

a point

If the centroid of an equilateral triangle is (1, 1) and its one vertex is (2, 2), then the equation of its circumcircle is _____________

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Answer :

x2 + y2 − 2x − 2y = 0

The equation of the incircle formed by the coordinate axes and the line 4x + 3y = 6 is_____________

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Answer :

4 (x2 + y2 − x − y) + 1 = 0

the circles x2 + y2 = 9 and x2 + y2 + 8y + 2c = 0 touch each other, then c is equal to _____________

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Answer :

15

If the circle x2 + y2 + 2ax + 6y + 9 = 0 touches x-axis, then the value of a is _____________

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Answer :

± 3

The equation of a circle with radius 4 and touching both the coordinate axes is _____________

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Answer :

xy± 8x ± 8y + 16 = 0

The equation of the circle passing through the origin which cuts off intercept of length 6 and 6 from the axes is _____________

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Answer :

x2 + y2 − 6x − 7y = 3√2 - 18

The circle x2 + y2 + 2gx + 2fy + c = 0 does not intersect x-axis, if g2 _____________

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Answer :

c

If (x, 3) and (3, 5) are the extremities of a diameter of a circle with centre at (2, y), then the values of x and y are _____________

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Answer :

x = 1, y = 4

Equation of the diameter of the circle x2 + y2 + 4x - 2y = 0 which passes through the origin is _____________

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Answer :

x + 2y = 0

The vertex of the parabola (y + k)2 = 8k (x − k) is _____________

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Answer :

(k, −k)

The equation of the parabola whose vertex is (k, 0) and the directrix has the equation y = 3k, is _____________

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Answer :

x2 − 2xy + y2 + 6kx + 10ky – 7k2 = 0

The locus of the points of trisection of the double ordinates of a parabola is a _____________

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Answer :

x2 − 2xy + y2 + 6kx + 10ky – 7k2 = 0

The equation of the parabola with focus (0, 0) and directrix x + y = 7 is _____________

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Answer :

x2 + y2 − 2xy + 14x + 14y − 49 = 0

In the parabola y2 = 4kx, the length of the chord passing through the vertex and inclined to the axis at π/4 is _____________

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Answer :

4√2k

The equation 9x2 + y2 + 6xy − 74x − 78y + 212 = 0 represents _____________

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Answer :

a parabola

Which points lie on the parabola x2 = ay?

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Answer :

x = aty = at2

Which points lie on the parabola x2 = 9ay?

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Answer :

x = 3aty = 3at2

Which points lie on the parabola 16x2 = 16ay?

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Answer :

x = aty = at2

Equation of the hyperbola whose vertices are (± 5, 0) and foci at (± 13, 0), is_____________

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Answer :

144x2 − 25y2 = 3600

If e1 and e2 are respectively the eccentricities of the ellipse x2/18 + y2/4 = 1 and the hyperbola x2/9 - y2/4 = 1, then the relation between e1 and e2 is_____________

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Answer :

e12 + e22 = 3

The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is_____________

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Answer :

8√2

The equation of the conic with focus at (1, −1) directrix along x − y + 1 = 0 and eccentricity √2 is_____________

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Answer :

2xy − 4x + 4y + 1 = 0

The eccentricity of the conic 25x2 − 144y2 = 3600 is_____________

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Answer :

13/12

Find the radius of each of the following circles :  x2 + y2 − 4x + 6y = 5

 

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2
Given:
x2 + y2 − 4x + 6y = 5

The given equation can be rewritten as follows:
(x−2)2+(y+3)2−4−9=5
⇒(x−2)2 + (y+3)2 =18

Thus, the radius = √18=3√2

Find the radius of each of the following circles :   (x + 5)2 + (y + 1)2 = 9

 

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2
Given:
(x + 5)2 + (y + 1)2 = 9
Thus, the radius = 3

Find the centre of each of the following circles :  x2 + y− x + 2y − 3 = 0.

 

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2
Given:
x2 + y2− x + 2y − 3=0

The given equation can be rewritten as follows:
(x− ½ )2+(y+1)2− ¼ −1−3=0
⇒(x− ½ )2 + (y+1)2 = 17/4
Thus, the centre is ( ½ ,−1)

Find the radius of each of the following circles :  (x − 1)2 + y2 = 4

 

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2
Given:
(x − 1)2 + y2 = 4
Thus, the radius is 2.

Find the centre of each of the following circles :  x2 + y2 − 4x + 6y = 5

 

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2
Given:
x2 + y2 − 4x + 6y = 5

The given equation can be rewritten as follows:
(x−2)2+(y+3)2−4−9=5
⇒(x−2)2 + (y+3)2 =18

Thus, the centre is (2, −3).

Find the centre of each of the following circles :  (x + 5)2 + (y + 1)2 = 9

 

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2
Given:
(x + 5)2 + (y + 1)2 = 9

Here, p = −5, q = −1

Thus, the centre is (-5, −1).

Find the centre of each of the following circles :  (x − 1)2 + y2 = 4

 

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2
Given:
(x − 1)2 + y2 = 4

Here, p = 1, q = 0 and a = 2

Thus, the centre is (1, 0)

Find the vertex of the following parabolas  :  4x2 + y = 0

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Answer :

Given:
  4x2 + y = 0
⇒-y/4=x2

On comparing the given equation with x2=−4ay:

4a=1/4

⇒a=1/16

∴ Vertex = (0, 0)

Find the vertex of the following parabolas  :  y2 = 8x

 

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Answer :

Given:
  y2 = 8x

On comparing the given equation with y2=4ax:
4a=8⇒a=2

∴ Vertex = (0, 0)

Find the radius of each of the following circles :  x2 + y− x + 2y − 3 = 0.

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Answer :

Let (p, q) be the centre of a circle with radius a.
Thus, its equation will be (x−p)2 + (y−q)2 = a2

Given:
x2 + y2− x + 2y − 3=0

The given equation can be rewritten as follows:
(x− ½ )2+(y+1)2− ¼ −1−3=0
⇒(x− ½ )2 + (y+1)2 = 17/4

Thus, the centre is ( ½ ,−1) and and the radius is √17/2.

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The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.

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