Chapter 11 Conic Sections Ex 11.3 Solutions
Question - 1 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 1 : -
Given:
The equation is x2/36+ y2/16 = 1
Here, the denominator of x2/36 isgreater than the denominator of y2/16.
So, the major axis is along the x-axis, while the minor axis isalong the y-axis.
On comparing the given equation with x2/a2 +y2/b2 =1, we get
a = 6 and b = 4.
c = √(a2 – b2)
= √(36-16)
= √20
= 2√5
Then,
The coordinates of the foci are (2√5, 0) and (-2√5, 0).
The coordinates of the vertices are (6, 0) and (-6, 0)
Length of major axis = 2a = 2 (6) = 12
Length of minor axis = 2b = 2 (4) = 8
Eccentricity, e = c/a =2√5/6 = √5/3
Length of latus rectum = 2b2/a= (2×16)/6 = 16/3
Question - 2 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 2 : -
Given:
The equation is x2/4 +y2/25 = 1
Here, the denominator of y2/25 isgreater than the denominator of x2/4.
So, the major axis is along the x-axis, while the minor axis isalong the y-axis.
On comparing the given equation with x2/a2 + y2/b2 = 1, we get
a = 5 and b = 2.
c = √(a2 – b2)
= √(25-4)
= √21
Then,
The coordinates of the foci are (0, √21) and (0, -√21).
The coordinates of the vertices are (0, 5) and (0, -5)
Length of major axis = 2a = 2 (5) = 10
Length of minor axis = 2b = 2 (2) = 4
Eccentricity, e = c/a = √21/5
Length of latus rectum = 2b2/a= (2×22)/5 = (2×4)/5 = 8/5
Question - 3 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 3 : -
Given:
The equation is x2/16+ y2/9 = 1 or x2/42 + y2/32 = 1
Here, the denominator of x2/16 isgreater than the denominator of y2/9.
So, the major axis is along the x-axis, while the minor axis isalong the y-axis.
On comparing the given equation with x2/a2 +y2/b2 =1, we get
a = 4 and b = 3.
c = √(a2 – b2)
= √(16-9)
= √7
Then,
The coordinates of the foci are (√7, 0) and (-√7, 0).
The coordinates of the vertices are (4, 0) and (-4, 0)
Length of major axis = 2a = 2 (4) = 8
Length of minor axis = 2b = 2 (3) = 6
Eccentricity, e = c/a = √7/4
Length of latus rectum = 2b2/a= (2×32)/4 = (2×9)/4 = 18/4 = 9/2
Question - 4 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 4 : -
Given:
The equation is x2/25+ y2/100 = 1
Here, the denominator of y2/100 isgreater than the denominator of x2/25.
So, the major axis is along the y-axis, while the minor axis isalong the x-axis.
On comparing the given equation with x2/b2 +y2/a2 =1, we get
b = 5 and a =10.
c = √(a2 – b2)
= √(100-25)
= √75
= 5√3
Then,
The coordinates of the foci are (0, 5√3) and (0, -5√3).
The coordinates of the vertices are (0, √10) and (0, -√10)
Length of major axis = 2a = 2 (10) = 20
Length of minor axis = 2b = 2 (5) = 10
Eccentricity, e = c/a = 5√3/10 = √3/2
Length of latus rectum = 2b2/a= (2×52)/10 = (2×25)/10 = 5
Question - 5 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 5 : -
Given:
The equation is x2/49+ y2/36 = 1
Here, the denominator of x2/49 isgreater than the denominator of y2/36.
So, the major axis is along the x-axis, while the minor axis isalong the y-axis.
On comparing the given equation with x2/a2 +y2/b2 =1, we get
b = 6 and a =7
c = √(a2 – b2)
= √(49-36)
= √13
Then,
The coordinates of the foci are (√13, 0) and (-√3, 0).
The coordinates of the vertices are (7, 0) and (-7, 0)
Length of major axis = 2a = 2 (7) = 14
Length of minor axis = 2b = 2 (6) = 12
Eccentricity, e = c/a = √13/7
Length of latus rectum = 2b2/a= (2×62)/7 = (2×36)/7 = 72/7
Question - 6 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 6 : -
Given:
The equation is x2/100+ y2/400 = 1
Here, the denominator of y2/400 isgreater than the denominator of x2/100.
So, the major axis is along the y-axis, while the minor axis isalong the x-axis.
On comparing the given equation with x2/b2 +y2/a2 =1, we get
b = 10 and a =20.
c = √(a2 – b2)
= √(400-100)
= √300
= 10√3
Then,
The coordinates of the foci are (0, 10√3) and (0, -10√3).
The coordinates of the vertices are (0, 20) and (0, -20)
Length of major axis = 2a = 2 (20) = 40
Length of minor axis = 2b = 2 (10) = 20
Eccentricity, e = c/a = 10√3/20 = √3/2
Length of latus rectum = 2b2/a= (2×102)/20 = (2×100)/20 = 10
Question - 7 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 7 : -
Given:
The equation is 36x2 +4y2 = 144 or x2/4 + y2/36= 1 or x2/22 +y2/62 =1
Here, the denominator of y2/62 is greater than the denominator of x2/22.
So, the major axis is along the y-axis, while the minor axis isalong the x-axis.
On comparing the given equation with x2/b2 +y2/a2 =1, we get
b = 2 and a = 6.
c = √(a2 – b2)
= √(36-4)
= √32
= 4√2
Then,
The coordinates of the foci are (0, 4√2) and (0, -4√2).
The coordinates of the vertices are (0, 6) and (0, -6)
Length of major axis = 2a = 2 (6) = 12
Length of minor axis = 2b = 2 (2) = 4
Eccentricity, e = c/a = 4√2/6 = 2√2/3
Length of latus rectum = 2b2/a= (2×22)/6 = (2×4)/6 = 4/3
Question - 8 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 8 : -
Given:
The equation is 16x2 +y2 = 16 or x2/1 + y2/16= 1 or x2/12 +y2/42 =1
Here, the denominator of y2/42 is greater than the denominator of x2/12.
So, the major axis is along the y-axis, while the minor axis isalong the x-axis.
On comparing the given equation with x2/b2 +y2/a2 =1, we get
b =1 and a =4.
c = √(a2 – b2)
= √(16-1)
= √15
Then,
The coordinates of the foci are (0, √15) and (0, -√15).
The coordinates of the vertices are (0, 4) and (0, -4)
Length of major axis = 2a = 2 (4) = 8
Length of minor axis = 2b = 2 (1) = 2
Eccentricity, e = c/a = √15/4
Length of latus rectum = 2b2/a= (2×12)/4 = 2/4 = ½
Question - 9 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Answer - 9 : -
Given:
The equation is 4x2 +9y2 = 36 or x2/9 + y2/4= 1 or x2/32 +y2/22 =1
Here, the denominator of x2/32 is greater than the denominator of y2/22.
So, the major axis is along the x-axis, while the minor axis isalong the y-axis.
On comparing the given equation with x2/a2 +y2/b2 =1, we get
a =3 and b =2.
c = √(a2 – b2)
= √(9-4)
= √5
Then,
The coordinates of the foci are (√5, 0) and (-√5, 0).
The coordinates of the vertices are (3, 0) and (-3, 0)
Length of major axis = 2a = 2 (3) = 6
Length of minor axis = 2b = 2 (2) = 4
Eccentricity, e = c/a = √5/3
Length of latus rectum = 2b2/a= (2×22)/3 = (2×4)/3 = 8/3
Question - 10 : - Find the equation for the ellipse that satisfies the given conditions:Vertices (± 5, 0), foci (± 4, 0)
Answer - 10 : -
Given:
Vertices (± 5, 0) and foci (± 4, 0)
Here, the vertices are on the x-axis.
So, the equation of the ellipse will be of the form x2/a2 +y2/b2 = 1, where ‘a’ is the semi-major axis.
Then, a = 5 and c = 4.
It is known that a2 = b2 + c2.
So, 52 = b2 + 42
25 = b2 + 16
b2 = 25 – 16
b = √9
= 3
∴ Theequation of the ellipse is x2/52 + y2/32 =1 or x2/25 + y2/9 = 1