# CBSE Solutions for Class 11 English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

The equations of the tangents to the ellipse 9x2 + 16y2 = 144 from the point (2, 3) are ___________

જવાબ : x + y = 5, y = 3

The latus-rectum of the conic 3x2 + 4y2 − 6x + 8y − 5 = 0 is ___________

જવાબ : 3

The eccentricity of the conic 9x2 + 25y2 = 225 is ___________

જવાબ : 4/5

The difference between the lengths of the major axis and the latus-rectum of an ellipse is ___________

જવાબ : 2ae2

The eccentricity of the ellipse, if the minor axis is equal to the distance between the foci, is ___________

જવાબ : 1/√2

The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latus-rectum, is ___________

જવાબ : √5-1/2

The eccentricity of the ellipse x2/a2 + y2/b2 =1 if its latus rectum is equal to one half of its minor axis, is___________

જવાબ : √3/2

The equation of the circle drawn with the two foci of x2/a2 + y2/b2 = 1 as the end-points of a diameter is___________

જવાબ : x2 + y2 = a2 − b2

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

જવાબ : (2/3, −1)

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

જવાબ : 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 _____________

જવાબ : x2 + y2 − 2x − 2y = 0

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

જવાબ : 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 _____________

જવાબ : 15

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

જવાબ : ± 3

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

જવાબ : 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 _____________

જવાબ : x2 + y2 − 6x − 7y = 3√2 - 18

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

જવાબ : < 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 _____________

જવાબ : x = 1, y = 4

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

જવાબ : x + 2y = 0

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

જવાબ : (k, −k)

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

જવાબ : x2 − 2xy + y2 + 6kx + 10ky – 7k2 = 0

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

જવાબ : x2 − 2xy + y2 + 6kx + 10ky – 7k2 = 0

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

જવાબ : x2 + y2 − 2xy + 14x + 14y − 49 = 0

The equation of the ellipse with focus (−1, 1), directrix x − y + 3 = 0 and eccentricity 1/2 is ___________

જવાબ : 7x2 + 2xy + 7y2 + 10x − 10y + 7 = 0

For the ellipse 12x2 + 4y2 + 24x − 16y + 25 = 0, centre is ___________

જવાબ : (−1, 2)

The locus of the point of intersection of the lines √3x-y-4√3λ=0 and √3λxy-4√3=0 is a hyperbola of eccentricity ___________

જવાબ : 2

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is ___________

જવાબ : 3 (x − 6)2 − (y −2)2 = 3

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

જવાબ : 4√2k

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

જવાબ : a parabola

Which points lie on the parabola x2 = ay?

જવાબ : x = aty = at2

Which points lie on the parabola x2 = 9ay?

જવાબ : x = 3aty = 3at2

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

જવાબ : x = aty = at2

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

જવાબ : 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_____________

જવાબ : 2 e12 + e22 = 3

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

જવાબ : 8√2

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

જવાબ : 2xy − 4x + 4y + 1 = 0

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

જવાબ : 13/12

A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA − PB = k (k ≠ 0), then the locus of P is_____________

જવાબ : a hyperbola

The eccentricity of the hyperbola which has latus-rectum ½ of its transverse axis, is_____________

જવાબ : √[3/2]

The eccentricity of the hyperbola x2 − 4y2 = 1 is _____________

જવાબ : √5/2

The difference of the focal distances of any point on the hyperbola is equal to _____________

જવાબ : length of the transverse axis

The foci of the hyperbola 25x2 − 144y2 = 3600 are _____________

જવાબ : (± 13, 0)

The eccentricity the hyperbola x=a2(t+1t), y=a2(t-1t) is ___________

જવાબ : √2

The foci of the hyperbola 2x2 − 3y2 = 5 are ___________

જવાબ : (±5/√6,0)

The latus-rectum of the hyperbola 16x2 − 9y2 = 144 is ___________

જવાબ : 32/3

The length of the straight line x − 3y = 1 intercepted by the hyperbola x2 − 4y2 = 1 is ___________

જવાબ : 6√2/5

The distance between the foci of a hyperbola is 8 and its eccentricity is √2, then equation of the hyperbola is_____________

જવાબ : x2 − y= 8

If e1 is the eccentricity of the conic 9x2 + 4y2 = 36 and e2 is the eccentricity of the conic 9x2 − 4y2 = 36, then e22 − e12 : _____________

જવાબ : 2 < e22 − e12 < 3

If the eccentricity of the hyperbola x2 − y2 sec2α = 5 is √3 times the eccentricity of the ellipse x2 sec2 α + y2 = 25, then α =___________

જવાબ : π/4

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is ___________

જવાબ : [(x-1)2]25/4 – [(y-4)2]75/4=1

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

જવાબ : 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

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

જવાબ : 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

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

જવાબ : 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

જવાબ : 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

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

જવાબ : 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

જવાબ : 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

જવાબ : 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 focus of the following parabolas  :  y2 = 8x + 8y

જવાબ : Given:
y2 = 8x + 8y
⇒(y−4)2=8(x+2)
Putting Y=y−4, X=x+2:
Y2=8X

On comparing the given equation with Y2=4aX:
4a=8

⇒a=2

∴ Focus = (X = a, Y = 0) = (x+2=2, y−4=0)=(x=0, y=4)

Find the focus of the following parabolas  :  y2 + 4x + 4y − 3 = 0

જવાબ : Given:
y2 + 4y + 4x −3 = 0

⇒(y+2)2− 4 + 4x – 3 = 0

⇒(y+2)2=−4 (x−7/4)
Let Y=y+2, X=x−7/4
Then, we have:
Y2=−4X

Comparing the given equation with Y2=−4aXY2=-4aX:
4a=4

⇒a=1

∴ Focus = (X = −a, Y = 0) = (x−74=−1, y+2=0)=(x=34, y=−2)

Find the focus of the following parabolas  :  y2 − 4y + 4x = 0

જવાબ : Given:
y2 − 4y + 4x = 0

⇒(y−2)2−4+4x=0

⇒(y−2)2=−4(x−1)

Let Y=y−2, X=x−1
Then, we have:
Y2=−4X

Comparing the given equation with Y2=−4aX:

4a=4

⇒a=1

∴ Focus = (X = −a, Y = 0) = (x−1=−1, y−2=0)=(x=0, y=2)

Find the focus of the following parabolas  :  y2 − 4y − 3x + 1 = 0

જવાબ : Given:
y2 − 4y − 3x + 1 = 0

⇒(y−2)2−4−3x+1=0

⇒(y−2)2 = 3(x+1)

⇒(y−2)2= 3(x−(−1))
Let Y=y−2, X=x+1
Then, we have:
Y2=3X
Comparing the given equation with Y2=4aX:
4a=3⇒a= ¾
∴ Focus = (X = a, Y = 0) = (x+1=34, y−2=0)=(x=−14, y=2)

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

જવાબ : Given:
4x2 + y = 0

⇒−y/4=x2

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

4a=1/4⇒

a=116

∴ Focus = (0, −a) = (0,−116)

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

જવાબ : Given:
y2 = 8x
On comparing the given equation with y2=4ax:
4a=8

⇒a=2

∴ Focus = (a, 0) = (2, 0)

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

જવાબ : Given:
y2 = 8x + 8y
⇒(y−4)2=8(x+2)

Putting Y=y−4, X=x+2:
Y2=8X

On comparing the given equation with Y2=4aX:
4a=8

⇒a=2

∴ Vertex = (X = 0, Y = 0) = (x=−2, y=4)x=-2, y=4

Find the vertex of the following parabolas  :  y2 + 4x + 4y − 3 = 0

જવાબ : Given:
y2 + 4y + 4x −3 = 0

⇒(y+2)2−4+4x−3=0

⇒(y+2)2=−4(x−7/4)

Let Y=y+2, X=x−7/4
Then, we have:
Y2=−4XY

Comparing the given equation with Y2=−4a:
4a=4

⇒a=1

∴ Vertex = (X = 0, Y = 0) = (x=74, y=−2)x=74, y=-2

Find the vertex of the following parabolas  :  y2 − 4y + 4x = 0

જવાબ : Given :

y2 − 4y + 4x = 0

⇒(y−2)2−4+4x=0

⇒(y−2)2=−4(x−1)
Let Y=y−2, X=x−1
Then, we have:
Y2=−4X

Comparing the given equation with Y2=−4aX:

4a=4

⇒a=1

∴ Vertex = (X = 0, Y = 0) = (x=1, y=2)x=1, y=2

Find the vertex of the following parabolas  :  y2 − 4y − 3x + 1 = 0

જવાબ : Given:
y2 − 4y − 3x + 1 = 0

⇒(y−2)2−4−3x+1=0

⇒(y−2)2=3(x+1)

⇒(y−2)2=3(x−(−1))

Let Y=y−2, X=x+1
Then, we have:
Y2=3X

Comparing the given equation with Y2=4aX:

4a=3

⇒a=34

∴ Vertex = (X = 0, Y = 0) = (x=−1, y=2)

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

જવાબ : 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

જવાબ : 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.

જવાબ : 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.