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NCERT Class 11 Mathematics Chapter 10 Conic Sections
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Conic Sections
Chapter – 10
| Exercise 10.1 |
In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2
Ans: The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
It is given that centre (h, k) = (0, 2) and radius (r) = 2
Therefore, the equation of the circle is
(x + 0)² + (y – 2)² = 2²
x² + y² + 4 – 4y = 4
x² + y² – 4y = 0.
2. centre (-2,3) and radius 4
Ans: The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
It is given that centre (h, k) = (- 2, 3) and radius (r) = 4
Therefore, the equation of the circle is
(x + 2)² + (y – 3)² = (4)²
x² + 4x + 4 + y² – 6y + 9 = 16
x² + y² + 4x – 6y – 3 = 0.
Ans: The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
Therefore, the equation of the circle is
144x² – 144x + 36 + 144y² – 72y + 9 – 1 = 0
144x² – 144x + 144y² – 72y + 44 = 0
36x² – 36x + 36y² – 18y + 11 = 0
36x² + 36y² – 36x – 18y + 11 = 0
4. centre (1,1) and radius √2
Ans: The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
It is given that centre (h, k) = (1, 1) and radius (r) = √2
Therefore, the equation of the circle is
(x – 1)² + (y – 1)² = (√2)²
x² – 2x + 1 + y² – 2y + 1 = 2
x² + y² – 2x – 2y = 0
5. centre (- a, – b) and radius √a² – b²)
Ans: The equation of a circle with centre (h, k) and radius r is given as
(x – h)² +(y – k)² = r²
It is given that centre (h, k) =(- a, – b) and radius (r) = √a² – b²
Therefore, the equation of the circle is
(x + a)² + (y + b)² = (√a² – b²)²
x² + 2ax + a² + y² + 2by + b² = a² – b²
x² + y² + 2ax + 2by + 2b² = 0
In each of the following Exercises 6 to 9, find the centre and radius of the circles.
6. (x + 5)² + (y – 3)² = 36
Ans: The equation of the given circle is (x + 5)² + (y – 3)² = 36
(x + 5)² + (y – 3)² = 36
⇒ {x – (- 5)}² + (y – 3)² = 6² , which is of the form (x – h)² + (y – k)² = r² ,where h = – 5 k = 3 and r = 6
Thus, the centre of the given circle is (-5, 3), while its radius is 6.
7. x² + y² – 4x – 8y – 45 = 0
Ans: The equation of the given circle is x² + y² – 4x – 8y – 45 = 0.
x² + y² – 4x – 8y – 45 = 0
⇒ (x² – 4x) + (y² – 8y) = 45
⇒{x² – 2(x)(2) + 2²} + {y² – 2(y)(4)+4²} – 4 – 16 = 45
⇒ (x – 2)² + (y – 4)² = 65
⇒ (x – 2)² + (y – 4)² = (√65)² , which is of the form (x – h)² + (y – k)² = r² where h = 2, k = 4 and r = √65
Thus, the centre of the given circle is (2, 4), while its radius is √65.
8. x² + y² – 8x + 10y – 12 = 0
Ans: The equation of the given circle is x² + y² – 8x + 10y – 12 = 0.
x² + y² – 8x + 10y – 12=0
⇒ (x² – 8x) + (y² + 10y) = 12
⇒{x² – 2(x)(4) + 4²} + {y² + 2(y)(5) + 5²} – 16 – 25 = 12
⇒ (x – 4)² + (y + 5)² = 53
⇒ (x – 4)² + {y – (- 5)}² = (√53)² ,which is of the form (x – h)² + (y – k)² = r² , where h = 4, k = – 5,
and r = √53
Thus, the centre of the given circle is ( 4 ,- 5) while its radius is √53
9. 2x² + 2y² – x = 0.
Ans: The equation of the given circle is 2x² + 2y² – x = 0 .
2x² + 2y² – x = 0
⇒ (2x² – x) + 2y² = 0
10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16
Ans: Let the equation of the required circle be (x – h)² +(y – k)² = r²
Since the circle passes through points (4, 1) and (6, 5),
( 4 – h)² +(1 – k)² =r² …(1)
(6 – h)² +(5 – k)² = r² … (2)
Since the centre (h, k) of the circle lies on line 4x + y = 16 ,
4h + k = 16 (3)
From equations (1) and (2), we obtain
( 4 – h)² +(1 – k)² =(6 – h)² +(5 – k)²
⇒ 16 – 8h + h² +1 – 2k + k²=36 – 12h + h² +25 – 10k+k²
⇒ 16 – 8h+1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 … (4)
On solving equations (3) and (4), we obtain h = 3 and k = 4
On substituting the values of h and k in equation (1), we obtain
(4 – 3)² +(1 – 4)² = r²
⇒ (1)² + (- 3)² = r²
⇒ 1 + 9 = r²
⇒ r² = 10
⇒ r = √10
Thus, the equation of the required circle is
(x – 3)² + (y – 4)² = (√10)²
x² – 6x + 9 + y² – 8y + 16 = 10
x² + y² – 6x – 8y + 15 = 0.
11. Find the equation of the circle passing through the points (2,3) and (-1,1) and whose centre is on the line x – 3y – 11 = 0
Ans: Let the equation of the required circle be (x – h)² + (y – k)² = r²
Since the circle passes through points (2, 3) and ( – 1,1),
( 2 – h)² +(3 – k)² =r² …(1)
( – 1 – h)² +(1 – k)² = r²…(2)
Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,
h – 3k =11…(3)
From equations (1) and (2), we obtain
(2 – h)² + (3 – k)² = (- 1 – h)² + (1- k)²
⇒ 4 – 4h + h² + 9 – 6k + k² = 1 + 2h + h² + 1 – 2k + k²
⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k
⇒ 6h + 4k =11…(4)
On solving equations (3) and (4), we obtain
On substituting the values of h and k in equation (1), we obtain
Thus, the equation of the required circle is
4x² – 28x + 49 + 4y² + 20y + 25 = 130
4x² + 4y² – 28x + 20y – 56 = 0
4(x² + y² – 7x + 5y – 14) = 0
x² + y² – 7x + 5y – 14 = 0
12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).
Ans: Let the equation of the required circle be (x – h)² +(y – k)² = r ²
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5
Now, the equation of the circle becomes (x – h)² + y² = 25 .
It is given that the circle passes through point (2, 3).
∴ (2 – h)² + 3² = 25
⇒ (2 – h)² = 25 – 9
⇒ (2 – h)² = 16
⇒ 2 – h = ±√16 = ± 4
If 2 – h = 4 then h = – 2
If 2 – h = – 4 then h = 6
When h = – 2, the equation of the circle becomes
(x + 2)² + y² = 25
x² + 4x + 4 + y² = 25
x² + y² +4x – 21=0
When h = 6 the equation of the circle becomes
(x – 6)² + y² = 25
x² – 12x + 36 + y² = 25
x² +y² – 12x + 11 = 0
13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.
Ans: Let the equation of the required circle be (x – h)² +(y – k)²= r²
Since the circle passes through (0, 0),
(0 – h)² +(0 – k)² = r²
⇒ h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k²
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b) Therefore,
(a – h)² + (0 – k)² = h² + k²…. (1)
(0 – h)² + (b – k)² = h² + k² …..(2)
From equation (1), we obtain
a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (b – 2k)=0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒h = a/2
From equation (2), we obtain
h² +b² – 2bk + k² = h² + k²
⇒ b² – 2bk = 0
⇒ b( b – 2k)=0
⇒ b = 0 or(b – 2k) = 0
However, b ≠ 0; hence, (b – 2k)=0⇒ k = b/2
Thus, the equation of the required circle is
⇒ 4x² – 4ax + a² + 4y² – 4by + b² = a² + b²
⇒ 4x² + 4y² – 4ax – 4by = 0
⇒ x² + y² – a x – by = 0
14. Find the equation of a circle with centre (2,2) and passes through the point (4,5)
Ans: The centre of the circle is given as (h, k) = (2, 2) .
Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).
∴ r = √(2 – 4)² + (2 – 5)² = √(- 2)² + (- 3)² = √4 + 9 = √13
Thus, the equation of the circle is
(x – h)² + (y – k)² = r²
(x – 2)² + (y – 2)² = (√13)²
x² – 4x + 4 + y² – 4y + 4 = 13
x² + y² – 4x – 4y – 5 = 0
15. Does the point (-2.5, 3.5) lie inside, outside or on the circle x² + y² = 25′
Ans: The equation of the given circle is x² + y² = 25
x² + y² = 25
⇒(x – 0)² +(y – 0)² = 5² , which is of the form (x – h)² + (y – k)² = r²,where h = 0, k = 0 and r = 5
∴ Centre = (0, 0) and radius = 5
Distance between point (- 2.5,3.5) and centre (0, 0)
= √(- 2.5 + 0)² + (3.5 + 0)²
= √6.25 + 12.25)
= √18.5
= 4.3 (approx.)<5
Since the distance between point ( – 2.5,3.5) and centre (0, 0) of the circle is less than the radius of the circle, point ( – 2.5,3.5) lies inside the circle.
| Exercise 10.2 |
In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
1. y² = 12x.
Ans: The given equation is y² = 12x
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax , we obtain
4a =12 ⇒ a=3
∴ Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y² the axis of the parabola is the x-axis.
Equation of directrix, x = – a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 x 3 = 12
2. x² = 6y.
Ans: The given equation is x² = 6y
Here, the coefficient of y is positive. Hence, the parabola opens upwards.
On comparing this equation with x² = 4ay, we obtain
∴ Coordinates of the focus = (0, a) =
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation of directrix,
Length of latus rectum = 4a= 6
3. y² = – 8x.
Ans: The given equation is y² = – 8x
Here, the coefficient of x is negative. Hence, the parabola opens towards the left.
On comparing this equation with y² = – 4ax we obtain
– 4a =-8 ⇒ a=2
∴ Coordinates of the focus = (- a, 0) = (- 2, 0)
Since the given equation involves y² the axis of the parabola is the x-axis.
Equation of directrix, x = a i.e., x = 2
Length of latus rectum = 4a= 8
4. x² = – 16y.
Ans: The given equation is x² = – 16y
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x² = – 4ay we obtain
-4a=-16 ⇒ a = 4
∴ Coordinates of the focus = (0, – a) = (0, – 4)
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation of directrix, y= a i.e., Y = 4
Length of latus rectum = 4a = 16
5. y² = 10x.
Ans: The given equation is y² = 10x
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax we obtain
Since the given equation involves y² the axis of the parabola is the x-axis.
Equation of directrix,
Length of latus rectum = 4a = 10
6. x² = – 9y.
Ans: The given equation is x² = – 9y
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x² = – 4ay, we obtain
∴ Coordinates of the focus =
Since the given equation involves x² the axis of the parabola is the y-axis.
Equation of directrix
Length of latus rectum = 4a= 9.
In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
7. Focus (6,0); directrix x = – 6
Ans: Focus (6, 0); directrix, x = – 6
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y² = 4ax or
y² = – 4ax
It is also seen that the directrix, x = – 6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.
Hence, the parabola is of the form y² = 4ax
Here, a = 6
Thus, the equation of the parabola is y² = 24x.
8. Focus (0,-3); directrix y = 3.
Ans: Focus = (0, -3); directrix y = 3
Since The focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x² = 4ay or
x² = – 4ay.
It is also seen that the directrix, y = 3 is above the x-axis, while the focus
(0, -3) is below the x-axis. Hence, the parabola is of the form x² = – 4ay
Here, a = 3
Thus, the equation of the parabola is x² = – 12y.
9. Vertex (0,0); focus (3,0).
Ans: Vertex (0, 0); focus (3, 0)
Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola,
while the equation of the parabola is of the form y² = 4ax
Since the focus is (3, 0), a = 3 .
Thus, the equation of the parabola is y² = 4 x 3 x x, i.e., y² = 12x
10. Vertex (0,0); focus (-2,0).
Ans: Vertex (0, 0) focus (-2, 0)
Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola,
while the equation of the parabola is of the form y² = – 4ax
Since the focus is (-2, 0), a = 2
Thus, the equation of the parabola is y² = – 4(2) x i.e., y² = – 8x.
11. Vertex (0,0) passing through (2,3) and axis is along x-axis.
Ans: Since the vertex is (0, 0) and the axis of the parabola isthe x-axis, the equation of the parabola is either of the
form y² = 4ax or y² = – 4ax
The parabola passes through point (2, 3), which lies in the first quadrant.
Therefore, the equation of the parabola is of the form y² = 4ax while point
(2, 3) must satisfy the equation y² = 4ax
Thus, the equation of the parabola is
12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis.
Ans: Since the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation of the parabola is either of the
formx²= 4ay or x² = – 4ay
The parabola passes through point (5, 2), which lies in the first quadrant.
Therefore, the equation of the parabola is of the form x² = 4ay while point
(5, 2) must satisfy the equation x² = 4ay
Thus, the equation of the parabola is
2x² = 25y
| Exercise 10.3 |
In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Ans: The given equation is
Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.
On comparing the given equation with
we obtain a = 6 and b = 4
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
Ans: The given equation is
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with
we obtain b = 2 and a = 5
∴ 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
Length of latus rectum
Ans: The given equation is
On comparing the given equation with
,we obtain a = 4 and b = 3
:. 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
Ans: The given equation is
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with
we obtain b = 5 and a = 10
:. c = √a² – b²) = √100 – 25 = √75 = 5√3
Therefore,
The coordinates of the foci are (0, ± 5√3)
The coordinates of the vertices are (0,±10).
Length of major axis = 2a = 20
Length of minor axis = 2b = 10
Ans: The Given equation is
Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.
On comparing the given equation with
, we obtain a = 7 and b = 6
∴ 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
Ans: The given equation is
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with
we obtain b = 10 and a = 20
∴ c = √a² – b²) = √400 – 100 = √300 = 10√3
Therefore,
The coordinates of the foci are (0, ±10√3)
The coordinates of the vertices are (0,±20)
Length of major axis = 2a = 40
Length of minor axis = 2b= 20
7. 36x² + 4y² = 144.
Ans: The given equation is 36x² + 4y² = 144
It can be written as
36x² + 4y² = 144
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing equation (1) with
, we obtain b = 2 and a = 6.
∴ c = √a² – b²) = √36 – 4 = √32 = 4√2
Therefore,
The coordinates of the foci are (0, ±4√2)
The coordinates of the vertices are (0, ± 6)
Length of major axis = 2a = 12
Length of minor axis = 2b = 4
Eccentricity,
Length of latus rectum
8. 16x² + y² = 16.
Ans: The given equation is 16x² + y² = 16
It can be written as
16x² + y² = 16
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis
On comparing equation (1) with
we obtain b = 1 and a = 4
∴ c = √a² – b²) = √16 – 1 = √15
Therefore,
The coordinates of the foci are (0, ±√15)
The coordinates of the vertices are (0, ± 4)
Length of major axis = 2a = 8
Length of minor axis = 2b = 2
9. 4x² + 9y² = 36.
Ans: The given equation is 4x² + 9y² = 36
It can be written as
4x² + 9y² = 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
we obtain a = 3 and b = 2
∴ c = √a² – b²) = √9 – 4 = √5
Therefore,
The coordinates of the foci are (±√5, 0)
The coordinates of the vertices are (± 3 ,0)
Length of major axis = 2a = 6
Length of minor axis = 2b = 4
In each of the following Exercises 10 to 20,find the equation for the ellipse that satisfies the given conditions:
10. Vertices (± 5,0), foci (±4, 0).
Ans: 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
where a is the semi-major axis.
Accordingly, a = 5 and c = 4
It is known that a² = b² + c²
∴ 5² = b² + 4²
⇒ 25 = b² + 16
⇒ b² = 25 – 16
⇒ b = √9 = 3
Thus, the equation of the ellipse is
11. Vertices (0, ±13), foci (0, ±5).
Ans: 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
,where a is the semi-major axis.
Accordingly, a = 13 and c = 5
Accordingly, a = 13 and c = 5
It is known that a² = b² + c²
∴ 13² = b² + 5²
⇒ 169 = b² + 25
⇒ b² = 169 – 25
⇒ b = √144 = 12
Thus, the equation of the ellipse is
12. Vertices (± 6, 0), foci (±4, 0).
Ans: 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
where a is the semi-major axis.
Accordingly, a = 6 c = 4
It is known that a² = b² + c²
∴ 6² = b² + 4²
⇒ 36 = b² + 16
⇒ b² = 36 – 16
⇒ b = √20
Thus, the equation of the ellipse is
13. Ends of major axis (± 3, 0), ends of minor axis ( ds of (0,±2)
Ans: 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
where a is the semi-major axis.
Accordingly, a = 3 and b = 2
Thus, the equation of the ellipse is
14. Ends of major axis (0,±√5), ends of minor axis (± 1,0)
Ans: 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
where a is the semi-major axis.
Accordingly, a = √5 and b = 1
Thus, the equation of the ellipse is
15. Length of major axis 26, foci (±5, 0)
Ans: 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
,where a is the semi-major axis
Accordingly, 2a =26 ⇒ a=13 and c = 5
It is known that a² = b² + c²
∴ 13² = b² + 5²
⇒ 169 = b² + 25
⇒ b² = 169 – 25
⇒ b = √144 = 12
Thus, the equation of the ellipse is
16. Length of minor axis 16, foci (0, ± 6)
Ans: 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 form
,where a is the semi-major axis.
Accordingly, 2b =16 ⇒ b=8 and c = 6
It is known that a² = b² + c²
∴ a² = 8² + 6² = 64 + 36 = 100
⇒ a = √100 = 10
Thus, the equation of the ellipse is
17. Foci (±3, 0), a = 4
Ans: 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
where a is the semi-major axis.
Accordingly, c = 3 and a = 4
It is known that a² = b² + c²
∴ 4² = b² + 3²
⇒16 = b² + 9
⇒ b² = 16 – 9 = 7
Thus, the equation of the ellipse is
18. b = 3 c = 4 centre at the origin; foci on the x axis.
Ans: 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
where a is the semi-major axis.
Accordingly, b = 3, c = 4
It is known that a² = b² + c²
∴ a² = 3² + 4² = 9 + 16 = 25
⇒ a = 5
Thus, the equation of the ellipse is
19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).
Ans: 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
Where, a is the semi-major axis
The ellipse passes through points (3, 2) and (1, 6). Hence,
On solving equations (2) and (3), we obtain b² = 10 and a² = 40
Thus, the equation of the ellipse is
20. Major axis on the x-axis and passes through the points (4,3) and (6,2).
Ans: Since the major axis is on the x-axis, the equation of the ellipse will be of the form
Where, a is the semi-major axis
The ellipse passes through points (4, 3) and (6, 2). Hence,
On solving equations (2) and (3), we obtain a² = 52 and b² = 13
Thus, the equation of the ellipse is
| Exercise 10.4 |
In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.
Ans: The given equation a
On comparing this equation with the standard equation of hyperbola i.e.
we obtain a = 4 and b = 3
We know that a² + b² = c²
∴ c² = 4² + 3² = 25
⇒ c = 5
Therefore,
The coordinates of the foci are ( ± 5 ,0).
The coordinates of the vertices are ( ± 4 ,0)
Length of latus rectum
Ans: The given equation is
On comparing this equation with the standard equation of hyperbola i.e
we obtain a = 3 and b = √27
We know that a² + b² = c²
∴ c² = 3² + (√27)² = 9 + 27 = 36
⇒ c = 6
Therefore,
The coordinates of the foci are (0, ± 6)
The coordinates of the vertices are (0, ± 3)
Length of latus rectum
3. 9y² – 4x² = 36.
Ans: The given equation is 9y² – 4x² = 36
It can be written as
9y – 4x² = 36
On comparing equation (1) with the standard equation of hyperbola i.e.
we obtain a = 2 and b = 3
We know that a² + b² = c²
∴ c² = 4 + 9 = 13
⇒ c = √13
Therefore,
The coordinates of the foci are (0, ± √13)
The coordinates of the vertices are (0, ± 2)
Length of latus rectum
4. 16x² – 9y² = 576.
Ans: The given equation is 16x² – 9y² = 576
It can be written as
16x² – 9y² = 576
On comparing equation (1) with the standard equation of hyperbola i.e.,
, we obtain a = 6 and b = 8
We know that a² + b² = c² .
∴ c² = 36 + 64 = 100
⇒ c = 10
Therefore,
The coordinates of the foci are ( ± 10 ,0)
The coordinates of the vertices are ( ± 6 ,0).
Length of latus rectum
5. 5y² – 9x² = 36.
Ans: The given equation is 5y² – 9x² = 36
On comparing equation (1) with the standard equation of hyperbola i.e.
2. We know that a² + b² = c²
Therefore, the coordinates of the foci are
The coordinates of the vertices are
Length of latus rectum
6. 49y² – 16x² = 784.
Ans: The given equation is 49y² – 16x² = 784
It can be written as
49y² – 16x² = 784
On comparing equation (1) with the standard equation of hyperbola i.e.,
we obtain a = 4 and b = 7
we know that a² + b² = c²
∴ c² = 16 + 49 = 65
⇒c = √65
The coordinates of the foci are (0, ± √65)
The coordinates of the vertices are (0, ± 4)
Length of latus rectum
In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.
7. Vertices (±2,0), foci (±3,0).
Ans: Vertices (± 2 ,0), foci ( ± 3 ,0)
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form
Since the vertices are ( ± 2 ,0), a = 2
Since the foci are ( ± 3 ,0) c = 3 .
We know that a² + b² = c² .
∴ 2² + b² = 3²
b² = 9 – 4 = 5
Thus, the equation of the hyperbola is
8. Vertices (0, ±5), foci (0,±8).
Ans: Vertices (0, ± 5) foci (0, ± 8)
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form
Since the vertices are (0, ± 5) a = 5
Since the foci are (0, ± 8) c = 8
We know that a² + b² = c²
∴ 5² + b² = 8²
b² = 64 – 25 = 39
Thus, the equation of the hyperbola is
9. Vertices (0, ±3), foci (0, ±5)
Ans: Vertices (0, ± 3) foci (0, ± 5)
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form
Since the vertices are (0, ± 3) a = 3 .
Since the foci are (0, ± 5) c = 5
We know that a² + b² = c²
∴ 3² + b² = 5²
⇒ b² =25 — 9=16
Thus, the equation of the hyperbola is
10. Foci (±5, 0), the transverse axis is of length 8.
Ans: Foci ( ± 5 ,0) the transverse axis is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form
Since the foci are ( ± 5 ,0) c = 5
Since the length of the transverse axis is 8, 2a =8 ⇒ a=4 .
We know that a² + b² = c²
∴ 4² + b² = 5²
⇒ b² =25 — 16=9
Thus, the equation of the hyperbola is
11. Foci (0, ±13), the conjugate axis is of length 24.
Ans: Foci (0, ± 13) the conjugate axis is of length 24.
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form
Since the foci are (0, ± 13) c = 13
Since the length of the conjugate axis is 24, 2b =24 ⇒ b=12
We know that a² + b² = c²
∴ a² + 12² = 13²
⇒ a² =169 – 144 = 25
Thus, the equation of the hyperbola is
12. Foci (± 3√5, 0) the latus rectum is of length 8.
Ans: Foci (± 3√5, 0), the latus rectum is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form
Since the foci are (± 3√5, 0), c = ± 3√5 .
Length of latus rectum = 8
⇒ b² = 4a
We know that a² + b² = c²
∴ a² + 4a = 45
⇒ a² + 4a – 45 = 0
⇒ a² +9a – 5a – 45 = 0
⇒ (a + 9) (a – 5)=0
⇒ a = – 9,5
Since a is non-negative, a = 5
∴ b² = 4a = 4 × 5 = 20
Thus, the equation of the hyperbola is
13. Foci (±4, 0), the latus rectum is of length 12.
Ans: Foci ( ± 4 ,0) the latus rectum is of length 12.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form
Since the foci are ( ± 4 ,0), c = 4 .
Length of latus rectum = 12
⇒ b² = 6a
We know that a² + b² = c²
∴ a² + 6a = 16
⇒ a² + 6a – 16=0
⇒ a² + 8a – 2a – 16=0
⇒ (a + 8) (a – 2)=0
⇒ a = – 8, 2
Since a is non-negative, a = 2 .
∴ b² = 6a = 6 × 2 = 12
Thus, the equation of the hyperbola is
14. vertices (± 7,0), e = 4/3.
Ans: Vertices ( ± 7 ,0),
Here, the vertices are on the x-axis
Therefore, the equation of the hyperbola is of the form
Since the vertices are ( ± 7 ,0), a = 7
It is given that
We know that a² + b² = c²
Thus, the equation of the hyperbola is
15. Foci (0, ± √10), passing through (2,3).
Ans: Foci (0, ±√10), passing through (2, 3)
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form
Since the foci are (0, ±√10),c = √10
We know that a² + b² = c²
∴ a² + b² = 10
⇒ b² =10 – a² …(1)
Since the hyperbola passes through point (2, 3),
From equations (1) and (2), we obtain
⇒ 9(10 – a²) – 4a² = a² (10 – a²)
⇒ 90 – 9a² – 4a² = 10a² – a⁴
⇒ a⁴ – 23a² + 90 = 0
⇒ a⁴ – 18a² – 5a² + 90 = 0
⇒ a² (a² – 18) – 5(a² – 18) = 0
⇒ (a² – 18)(a² – 5) = 0
⇒ a² = 18 or 5
In hyperbola, c > a i.e., c² > a²
∴ a² = 5
⇒ b² = 10 – a² = 10 – 5 = 5
Thus, the equation of the hyperbola is
| Miscellaneous Exercise on Chapter 10 |
1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Ans: The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis.
This can be diagrammatically represented as
The equation of the parabola is of the form y² = 4ax (as it is opening to the right).
Since the parabola passes through point A (10, 5), 10² = 4a(5)
⇒ 100 = 20a
Therefore, the focus of the parabola is (a, 0) = (5, 0) , which is the mid – point of the diameter.
Hence, the focus of the reflector is at the mid – point of the diameter.
2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?
Ans: The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the positive y – axis.
This can be diagrammatically represented as
The equation of the parabola is of the form x² = 4ay (as it is opening upwards).
It can be clearly seen that the parabola passes through point
Therefore, the arch is in the form of a parabola whose equation is
Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.
3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Ans: The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola,
while its vertical axis is taken along the positive y-axis. This can be diagrammatically represented as
Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.
DF is the supporting wire attached to the roadway, 18 m from the middle.
Here, AB = 30m OC = 6 m, and
The equation of the parabola is of the form x² = 4ay (as it is opening upwards).
The coordinates of point A are (50,30 — 6)= (50, 24) .
Since A (50, 24) is a point on the parabola,
(50)² = 4a(24)
The x-coordinate of point D is 18.
Hence, at x = 18
6(18)² = 625y
⇒ y = 3.11 (approx)
∴ DE = 3.11m
DF = DE + EF = 3.11m + 6m = 9.11m
Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately 9.11 m.
4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.
Ans: Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.
The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis. Hence, the semi-ellipse can be diagrammatically represented as
The equation of the semi-ellipse will be of the form
where a is the semi-major axis
Accordingly, 2a =8 ⇒ a = 4
b = 2
Therefore, the equation of the semi-ellipse is
Let A be a point on the major axis such that AB = 1.5m
Draw AC ⏊ ОВ.
OA = (4 — 1.5) m = 2.5m
The x-coordinate of point C is 2.5.
On substituting the value of x with 2.5 in equation (1), we obtain
⇒ y² = 2.4375
⇒ y = 1.56 (approx.)
∴ AC = 1.56m
Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.
5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.
Ans: Let AB be the rod making an angle θ with OX and P (x, y) be the point on it such that AP = 3 cm.
Then, PB = AB – AP = (12 – 3) cm = 9 cm [AB = 12 cm]
From P, draw PQ ⏊ OY and PR ⏊ OX.
Since, sin² θ + cos³ θ = 1
Thus, the equation of the locus of point P on the rod is
6. Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of its latus rectum.
Ans: The given parabola is x² = 12y
On comparing this equation with x² = 4ay we obtain 4a =12 ⇒ a=3
∴ The coordinates of foci are S (0, a) = S (0, 3)
Let AB be the latus rectum of the given parabola.
The given parabola can be roughly drawn as
At y = 3 , x² =12(3) ⇒ x² =36 ⇒ x = ± 6
∴ The coordinates of A are ( – 6,3) , while the coordinates of B are (6, 3).
Therefore, the vertices of triangle OAB are Ο (0, 0), A ( – 6,3) , and B (6, 3).
Thus, the required area of the triangle is 18 unit².
7. A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.
Ans: Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.
We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.
Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.
Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as
The equation of the ellipse will be of the form
where a is the semi-major axis
Accordingly, 2a =10 ⇒ a=5
Distance between the foci (2c) = 8
⇒ c = 4
On using the relation c = √a² – b² , we obtain
4 = √25 – b²
⇒ 16 = 25 – b²
⇒ b² = 25 – 16 = 9
⇒ b = 3
Thus, the equation of the path traced by the man is
8. An equilateral triangle is inscribed in the parabola y² = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Ans: Let OAB be the equilateral triangle inscribed in parabola y² = 4ax
Let AB intersect the x-axis at point C.
Let OC = k
From the equation of the given parabola, we have y² =4ak ⇒ y = ± 2√ak
∴ The respective coordinates of points A and B are (k, 2√ak) and (k, – 2√ak)
AB = CA + CB = 2√ak + 2√ak = 4√ak
Since OAB is an equilateral triangle, OA² = AB²
∴ k² + (2√ak)² = (4√ak)²
⇒ k² + 4ak = 16ak
⇒ k² = 12ak
⇒ k = 12a
∴ AB = 4√ak = 4√a × 12a = 4√12a² = 8√3a
Thus, the side of the equilateral triangle inscribed in parabola y² = 4ax is 8√3a
