Class 11 - Mathematics
Conic Sections - Exercise 11.3
Top Block 1
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 x2/36 + y2/16 = 1.
Answer:
The given equation is x2/36 + y2/16 = 1 or or x2/62 + y2/42 = 1.
Here, the denominator of x2/36 is greater than the denominator of y2/16.
Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.
On comparing the given equation with x2/a2 + y2/b2 = 1
we obtain a = 6 and b = 4.
Now, c = √(a2 – b2) = √(36 – 16) = √20 = 2√5
Therefore,
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 = 12
Length of minor axis = 2b = 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 x2/4 + y2/25 = 1.
Answer:
The given equation is x2/4 + y2/25 = 1 or or x2/22 + y2/52 = 1.
Here, the denominator of y2/25 is greater than the denominator of x2/4.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with x2/b2 + y2/a2 = 1,
we get b = 2 and a = 5.
Now, c = √(a2 – b2) = √(25 – 4) = √21
Therefore,
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 = 10
Length of minor axis = 2b = 4
Eccentricity, e = c/a = √21/5
Length of latus rectum = 2b2/a = (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 x2/16 + y2/9 = 1.
Answer:
The given equation is x2/16 + y2/9 = 1 or x2/42 + y2/32 = 1.
Here, the denominator of x2/16 is greater than the denominator of y2/9.
Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.
On comparing the given equation with x2/a2 + y2/b2 = 1,
we obtain a = 4 and b = 3.
Now, c = √(a2 – b2) = √(16 – 9) = √7
Therefore,
The coordinates of the foci are (±√7, 0)
The coordinates of the vertices are (±4, 0)
Length of major axis = 2a = 8
Length of minor axis = 2b = 6
Eccetricity, e = c/a = √7/4
Length of latus rectum = 2b2/a = (2 * 9)/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 x2/25 + y2/100 = 1.
Answer:
The given equation is x2/25 + y2/100 = 1 or or x2/52 + y2/102 = 1.
Here, the denominator of y2/100 is greater than the denominator of x2/25.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with x2/b2 + y2/a2 = 1,
we get b = 5 and a = 10.
Now, c = √(a2 – b2) = √(100 – 25) = √75 = 5√3
Therefore,
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 = 20
Length of minor axis = 2b = 10
Eccentricity, e = c/a = 5√3/10 = √3/2
Length of latus rectum = 2b2/a = (2 * 25)/10 = 25/5 = 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 x2/49 + y2/36 = 1.
Answer:
The given equation is x2/49 + y2/36 = 1 or x2/72 + y2/62 = 1.
Here, the denominator of x2/49 is greater than the denominator of y2/36.
Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.
On comparing the given equation with x2/a2 + y2/b2 = 1,
we obtain a = 7 and b = 6.
Now, c = √(a2 – b2) = √(49 – 36) = √13
Therefore,
The coordinates of the foci are (±√13, 0)
The coordinates of the vertices are (±7, 0)
Length of major axis = 2a = 14
Length of minor axis = 2b = 12
Eccetricity, e = c/a = √13/7
Length of latus rectum = 2b2/a = (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 x2/100 + y2/400 = 1.
Answer:
The given equation is x2/100 + y2/400 = 1 or or x2/102 + y2/202 = 1.
Here, the denominator of y2/400 is greater than the denominator of x2/100.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with x2/b2 + y2/a2 = 1,
we get b = 10 and a = 20.
Now, c = √(a2 – b2) = √(400 – 100) = √300 = 10√3
Therefore,
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 = 40
Length of minor axis = 2b = 20
Eccentricity, e = c/a = 10√3/20 = √3/2
Length of latus rectum = 2b2/a = (2 * 100)/20 = 2 * 5 = 10
Mddle block 1
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 36x2 + 4y2 = 144.
Answer:
The given equation is 36x2 + 4y2 = 144
36x2/144 + 4y2/144 = 1
x2/4 + y2/36 = 1
x2/22 + y2/62 = 1.
Here, the denominator of y2/36 is greater than the denominator of x2/4.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with x2/b2 + y2/a2 = 1,
we get b = 2 and a = 6
Now, c = √(a2 – b2) = √(36 – 4) = √32 = 4√2
Therefore,
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 = 12
Length of minor axis = 2b = 4
Eccentricity, e = c/a = 4√2/6 = 2√2/3
Length of latus rectum = 2b2/a = (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 16x2 + y2 = 16.
Answer:
The given equation is 16x2 + y2 = 16
16x2/16 + y2/16 = 1
x2/1 + y2/16 = 1
x2/12 + y2/42 = 1.
Here, the denominator of y2/4 is greater than the denominator of x2/1.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with x2/b2 + y2/a2 = 1,
we get b = 1 and a = 4
Now, c = √(a2 – b2) = √(16 – 1) = √15
Therefore,
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 = 8
Length of minor axis = 2b = 2
Eccentricity, e = c/a = √15/4
Length of latus rectum = 2b2/a = (2 * 1)/4 = 1/2
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 4x2 + 9y2 = 36.
Answer:
The given equation is 4x2 + 9y2 = 36
4x2/36 + 9y2/36 = 1
x2/9 + y2/4 = 1
x2/32 + y2/22 = 1.
Here, the denominator of x2/9 is greater than the denominator of y2/4.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with x2/a2 + y2/b2 = 1,
we get b = 2 and a = 3
Now, c = √(a2 – b2) = √(9 – 4) = √5
Therefore,
The coordinates of the foci are (0, √5) and (0, -√5)
The coordinates of the vertices are (±3, 4)
Length of major axis = 2a = 6
Length of minor axis = 2b = 4
Eccentricity, e = c/a = √5/3
Length of latus rectum = 2b2/a = (2 * 3)/3 = 8/3
Question 10:
Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci (±4, 0).
Answer:
Given, Vertices (±5, 0), foci (±4, 0)
Here, the vertices are on the x-axis.
Therefore, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where a is the
semi-major axis.
Accordingly, a = 5 and c = 4.
We know that
a2 = b2 + c2
=> 52 = b2 + 42
=> b2 = 25 – 16
=> b2 = 9
=> b = 3
Thus, the equation of the ellipse is x2/52 + y2/32 = 1 or x2/25 + y2/9 = 1
Question 11:
Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13), foci (0, ±5).
Answer:
Given, Vertices (0, ±13), foci (0, ±5).
Here, the vertices are on the y-axis.
Therefore, the equation of the ellipse will be of the form x2/b2 + y2/a2 = 1, where a is the
semi-major axis.
Accordingly, a = 13 and c = 5.
We know that
a2 = b2 + c2
=> 132 = b2 + 52
=> b2 = 169 – 25
=> b2 = 144
=> b = 12
Thus, the equation of the ellipse is x2/122 + y2/132 = 1 or x2/144 + y2/169 = 1
Question 12:
Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci (±4, 0).
Answer:
Given, Vertices (±6, 0), foci (±4, 0)
Here, the vertices are on the x-axis.
Therefore, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where a is the
semi-major axis.
Accordingly, a = 6 and c = 4.
We know that
a2 = b2 + c2
=> 62 = b2 + 42
=> b2 = 36 – 16
=> b2 = 20
=> b = √20
Thus, the equation of the ellipse is x2/62 + y2/(√20)2 = 1 or x2/36 + y2/20 = 1
Question 13:
Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (±3, 0), ends of minor axis (0, ±2).
Answer:
Given, Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Here, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where a is the
semi-major axis.
Accordingly, a = 3 and b = 2.
Thus, the equation of the ellipse is x2/32 + y2/22 = 1 or x2/9 + y2/4 = 1
Question 14:
Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ±√5), ends of minor axis (±1, 0)
Answer:
Given, Ends of major axis (0, ±√5), ends of minor axis (±1, 0)
Here, the major axis is along the y-axis.
Therefore, the equation of the ellipse will be of the form x2/b2 + y2/a2 = 1, where a is the
semi-major axis.
Accordingly, a = √5 and b = 1.
Thus, the equation of the ellipse is x2/12 + y2/(√5)2 = 1 or x2/1 + y2/5 = 1
Question 15:
Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (±5, 0).
Answer:
Length of major axis = 26; foci = (±5, 0).
Since the foci are on the x-axis, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where a is the
semi-major axis.
Accordingly, 2a = 26
=> a = 13 and c = 5.
We know that
a2 = b2 + c2
=> 132 = b2 + 52
=> b2 = 169 – 25
=> b2 = 144
=> b = 12
Thus, the equation of the ellipse is x2/132 + y2/122 = 1 or x2/169 + y2/144 = 1
Question 16:
Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ±6).
Answer:
Length of minor axis 16, foci (0, ±6).
Since the foci are on the y-axis, the major axis is along the y-axis.
Therefore, the equation of the ellipse will be of the form x2/b2 + y2/a2 = 1, where a is the
semi-major axis.
Accordingly, 2b = 16
=> b = 8 and c = 6.
We know that
a2 = b2 + c2
=> a2 = 82 + 62
=> a2 = 64 + 36
=> a2 = 100
=> a = 10
Thus, the equation of the ellipse is x2/82 + y2/102 = 1 or x2/64 + y2/100 = 1
Question 17:
Find the equation for the ellipse that satisfies the given conditions: foci (±3, 0), a = 4
Answer:
Given, foci (±3, 0), a = 4.
Since the foci are on the x-axis, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where a is the
semi-major axis.
Accordingly, a = 4 and c = 3.
We know that
a2 = b2 + c2
=> 42 = b2 + 32
=> b2 = 16 – 9 => b2 = 7
Thus, the equation of the ellipse is x2/16 + y2/7 = 1
Question 18:
Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the x axis.
Answer:
It is given that b = 3, c = 4, centre at the origin; foci on the x axis.
Since the foci are on the x-axis, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where a is the
semi-major axis.
Accordingly, b = 3, c = 4.
We know that
a2 = b2 + c2
=> a2 = 32 + 42
=> a2 = 9 + 16
=> a2 = 25
=> a = 5
Thus, the equation of the ellipse is x2/52 + y2/32 = 1 or x2/25 + y2/9 = 1
Question 19:
Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).
Answer:
Since the centre is at (0, 0) and the major axis is on the y-axis, the equation of the ellipse will
be of the form
x2/b2 + y2/a2 = 1 ………….1
where a is the semi-major axis.
The ellipse passes through points (3, 2) and (1, 6). Hence,
9/b2 + 4/a2 = 1 ………….2
1/b2 + 36/a2 = 1 ………….3
On solving equations 2 and 3, we get
b2 = 10 and a2 = 40
Thus, the equation of the ellipse is x2/10 + y2/40 = 1 or 4x2 + y2 = 40
Question 20:
Find the equation for the ellipse that satisfies the given conditions: Major axis on the x -axis and passes through the points (4, 3) and (6, 2).
Answer:
Since the major axis is on the x-axis, the equation of the ellipse will be of the form
x2/b2 + y2/a2 = 1 ………….1
where a is the semi-major axis.
The ellipse passes through points (4, 3) and (6, 2). Hence,
16/b2 + 9/a2 = 1 ………….2
36/b2 + 4/a2 = 1 ………….3
On solving equations 2 and 3, we get
a2 = 52 and b2 = 13.
Thus, the equation of the ellipse is x2/52 + y2/13 = 1 or x2 + 4y2 = 52
Bottom Block 3
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