જવાબ : x + y = 5, y = 3
જવાબ : 3
જવાબ : 4/5
જવાબ : 2ae2
જવાબ : 1/√2
જવાબ : √5-1/2
જવાબ : √3/2
જવાબ : x2 + y2 = a2 − b2
જવાબ : (2/3, −1)
જવાબ : a point
જવાબ : x2 + y2 − 2x − 2y = 0
જવાબ : 4 (x2 + y2 − x − y) + 1 = 0
જવાબ : 15
જવાબ : ± 3
જવાબ : x2 + y2 ± 8x ± 8y + 16 = 0
જવાબ : x2 + y2 − 6x − 7y = 3√2 - 18
જવાબ : < c
જવાબ : x = 1, y = 4
જવાબ : x + 2y = 0
જવાબ : (k, −k)
જવાબ : x2 − 2xy + y2 + 6kx + 10ky – 7k2 = 0
જવાબ : x2 − 2xy + y2 + 6kx + 10ky – 7k2 = 0
જવાબ : x2 + y2 − 2xy + 14x + 14y − 49 = 0
જવાબ : 7x2 + 2xy + 7y2 + 10x − 10y + 7 = 0
જવાબ : (−1, 2)
જવાબ : 2
જવાબ : 3 (x − 6)2 − (y −2)2 = 3
જવાબ : 4√2k
જવાબ : a parabola
જવાબ : x = at, y = at2
જવાબ : x = 3at, y = 3at2
જવાબ : x = at, y = at2
જવાબ : 144x2 − 25y2 = 3600
જવાબ : 2 e12 + e22 = 3
જવાબ : 8√2
જવાબ : 2xy − 4x + 4y + 1 = 0
જવાબ : 13/12
જવાબ : a hyperbola
જવાબ : √[3/2]
જવાબ : √5/2
જવાબ : length of the transverse axis
જવાબ : (± 13, 0)
જવાબ : √2
જવાબ : (±5/√6,0)
જવાબ : 32/3
જવાબ : 6√2/5
જવાબ : x2 − y2 = 8
જવાબ : 2 < e22 − e12 < 3
જવાબ : π/4
જવાબ : [(x-1)2]25/4 – [(y-4)2]75/4=1
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 radius = √18=3√2
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
Thus, the radius = 3
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)
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
Thus, the radius is 2.
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).
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).
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)
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
જવાબ : Given:
y2 + 4y + 4x −3 = 0
⇒(y+2)2− 4 + 4x – 3 = 0
જવાબ : Given:
y2 − 4y + 4x = 0
⇒(y−2)2−4+4x=0
જવાબ : Given:
y2 − 4y − 3x + 1 = 0
⇒(y−2)2−4−3x+1=0
જવાબ : Given:
4x2 + y = 0
⇒−y/4=x2
On comparing the given equation with x2=−4ay:
4a=1/4⇒
જવાબ : Given:
y2 = 8x
On comparing the given equation with y2=4ax:
4a=8
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
જવાબ : Given:
y2 + 4y + 4x −3 = 0
⇒(y+2)2−4+4x−3=0
જવાબ : Given :
y2 − 4y + 4x = 0જવાબ : Given:
y2 − 4y − 3x + 1 = 0
⇒(y−2)2−4−3x+1=0
જવાબ : Given:
4x2 + y = 0
⇒-y/4=x2
On comparing the given equation with x2=−4ay:
4a=1/4
જવાબ : Given:
y2 = 8x
On comparing the given equation with y2=4ax:
4a=8⇒a=2
∴ Vertex = (0, 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
Math
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.