Chapter – 11: Conic Sections
EXERCISE 11.1
In each of the following Exercise 1 to 5, find the equation of the circle with
1. Centre (0, 2) and radius 2
Solution: –
Given:
Centre (0, 2) and radius 2
Let us consider the equation of a circle with centre (h, k) and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, centre (h, k) = (0, 2) and radius (r) = 2
The equation of the circle is
(x – 0)2 + (y – 2)2 = 22
x2 + y2 + 4 – 4y = 4
x2 + y2 – 4y = 0
∴ The equation of the circle is x2 + y2 – 4y = 0
2. Centre (–2, 3) and radius 4
Solution: –
Given:
Centre (-2, 3) and radius 4
Let us consider the equation of a circle with centre (h, k) and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, centre (h, k) = (-2, 3) and radius (r) = 4
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
∴ The equation of the circle is x2 + y2 + 4x – 6y – 3 = 0
3. Centre (1/2, 1/4) and radius (1/12)
Solution: –
Given:
Centre (1/2, 1/4) and radius 1/12
Let us consider the equation of a circle with centre (h, k) and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, centre (h, k) = (1/2, 1/4) and radius (r) = 1/12
The equation of the circle is
(x – 1/2)2 + (y – 1/4)2 = (1/12)2
x2 – x + ¼ + y2 – y/2 + 1/16 = 1/144
x2 – x + ¼ + y2 – y/2 + 1/16 = 1/144
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
∴ The equation of the circle is 36x2 + 36y2 – 36x – 18y + 11= 0
4. Centre (1, 1) and radius √2
Solution: –
Given:
Centre (1, 1) and radius √2
Let us consider the equation of a circle with centre (h, k) and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, centre (h, k) = (1, 1) and radius (r) = √2
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
∴ The equation of the circle is x2 + y2 – 2x -2y = 0
5. Centre (–a, –b) and radius √(a2 – b2)
Solution: –
Given:
Centre (-a, -b) and radius √(a2 – b2)
Let us consider the equation of a circle with centre (h, k) and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, centre (h, k) = (-a, -b) and radius (r) = √(a2 – b2)
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
∴ The equation of the circle is x2 + y2 +2ax + 2by + 2b2 = 0
In each of the following Exercise 6 to 9, find the centre and radius of the circles.
6. (x + 5)2 + (y – 3)2 = 36
Solution: –
Given:
The equation of the given circle is (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
∴ The centre of the given circle is (-5, 3) and its radius is 6.
7. x2 + y2 – 4x – 8y – 45 = 0
Solution: –
Given:
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 form (x-h)2 +(y-k)2 = r2]
Where h = 2, K = 4 and r = √65
∴ The centre of the given circle is (2, 4) and its radius is √65.
8. x2 + y2 – 8x + 10y – 12 = 0
Solution: –
Given:
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 form (x-h)2 +(y-k)2= r2]
Where h = 4, K= -5 and r = √53
∴ The centre of the given circle is (4, -5) and its radius is √53.
9. 2x2 + 2y2 – x = 0
Solution: –
The equation of the given of the circle is 2x2 + 2y2 –x = 0.
2x2 + 2y2 –x = 0
(2x2 + x) + 2y2 = 0
(x2 – 2 (x) (1/4) + (1/4)2) + y2 – (1/4)2 = 0
(x – 1/4)2 + (y – 0)2 = (1/4)2 [which is form (x-h)2 +(y-k)2= r2]
Where, h = ¼, K = 0, and r = ¼
∴ The center of the given circle is (1/4, 0) and its radius is 1/4.
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.
Solution: –
Let us consider the equation of the required circle be (x – h)2+ (y – k)2 = r2
We know that the circle passes through points (4,1) and (6,5)
So,
(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 the equation (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+15 – 10k + k2
16 – 8h +1 -2k + 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
r = √10
so now, (x – 3)2 + (y – 4)2 = (√10)2
x2 – 6x + 9 + y2 – 8y + 16 =10
x2 + y2 – 6x – 8y + 15 = 0
∴ The equation of the required circle is 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.
Solution: –
Let us consider the equation of the required circle be (x – h)2 + (y – k)2 = r2
We know that 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 the equation (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)
Now let us multiply equation (3) by 6 and subtract it from equation (4) to get,
6h+ 4k – 6(h-3k) = 11 – 66
6h + 4k – 6h + 18k = 11 – 66
22 k = – 55
K = -5/2
Substitute this value of K in equation (4) to get,
6h + 4(-5/2) = 11
6h – 10 = 11
6h = 21
h = 21/6
h = 7/2
We obtain h = 7/2and k = -5/2
On substituting the values of h and k in equation (1), we get
(2 – 7/2)2 + (3 + 5/2)2 = r2
[(4-7)/2]2 + [(6+5)/2]2 = r2
(-3/2)2 + (11/2)2 = r2
9/4 + 121/4 = r2
130/4 = r2
The equation of the required circle is
(x – 7/2)2 + (y + 5/2)2 = 130/4
[(2x-7)/2]2 + [(2y+5)/2]2 = 130/4
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
∴ The equation of the required circle is 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).
Solution: –
Let us consider the equation of the required circle be (x – h)2+ (y – k)2 = r2
We know that the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
So now, the equation of the circle is (x – h)2 + y2 = 25.
It is given that the circle passes through the point (2, 3) so the point will satisfy the equation of the circle.
(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
Then, when h = -2, the equation of the circle becomes
(x + 2)2 + y2 = 25
x2 + 12x + 36 + 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
∴ The equation of the required circle is x2 + y2 + 4x – 21 = 0 and 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.
Solution: –
Let us consider the equation of the required circle be (x – h)2+ (y – k)2 =r2
We know that the circle passes through (0, 0),
So, (0 – h)2+ (0 – k)2 = r2
h2 + k2 = r2
Now, The equation of the circle is (x – h)2 + (y – k)2 = h2 + k2.
It is given that the circle intercepts a and b on the coordinate axes.
i.e., the circle passes through points (a, 0) and (0, b).
So, (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 (a -2h) = 0
However, a ≠ 0; hence, (a -2h) = 0
h = a/2
From equation (2), we obtain
h2 – 2bk + k2 + b2= h2 +k2
b2 – 2bk = 0
b(b– 2k) = 0
b= 0 or (b-2k) =0
However, a ≠ 0; hence, (b -2k) = 0
k =b/2
So, the equation is
(x – a/2)2 + (y – b/2)2 = (a/2)2 + (b/2)2
[(2x-a)/2]2 + [(2y-b)/2]2 = (a2 + b2)/4
4x2 – 4ax + a2 +4y2 – 4by + b2 = a2 + b2
4x2 + 4y2 -4ax – 4by = 0
4(x2 +y2 -7x + 5y – 14) = 0
x2 + y2 – ax – by = 0
∴ The equation of the required circle is x2 + y2 – ax – by = 0
14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).
Solution: –
Given:
The centre of the circle is given as (h, k) = (2,2)
We know that 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
The equation of the circle is given as
(x– h)2+ (y – k)2 = r2
(x –h)2 + (y – k)2 = (√13)2
(x –2)2 + (y – 2)2 = (√13)2
x2 – 4x + 4 + y2 – 4y + 4 = 13
x2 + y2 – 4x – 4y = 5
∴ The equation of the required circle is x2 + y2 – 4x – 4y = 5
15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Solution: –
Given:
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.
So the distance between point (-2.5, 3.5) and the centre (0,0) is
= √[(-2.5 – 0)2 + (-3.5 – 0)2]
= √(6.25 + 12.25)
= √18.5
= 4.3 [which is < 5]
Since, the distance between point (-2.5, -3.5) and the centre (0, 0) of the circle is less than the radius of the circle, point (-2.5, -3.5) lies inside the circle.
EXERCISE 11.2
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
Solution: –
Given:
The equation is y2 = 12x
Here we know that the coefficient of x is positive.
So, the parabola opens towards the right.
On comparing this equation with y2 = 4ax, we get,
4a = 12
a = 3
Thus, the co-ordinates of the focus = (a, 0) = (3, 0)
Since, the given equation involves y2, the axis of the parabola is the x-axis.
∴ The equation of directrix, x = -a, then,
x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12
2. x2 = 6y
Solution: –
Given:
The equation is x2 = 6y
Here we know that the coefficient of y is positive.
So, the parabola opens upwards.
On comparing this equation with x2 = 4ay, we get,
4a = 6
a = 6/4
= 3/2
Thus, the co-ordinates of the focus = (0,a) = (0, 3/2)
Since, the given equation involves x2, the axis of the parabola is the y-axis.
∴ The equation of directrix, y =-a, then,
y = -3/2
Length of latus rectum = 4a = 4(3/2) = 6
3. y2 = – 8x
Solution: –
Given:
The equation is y2 = -8x
Here we know that the coefficient of x is negative.
So, the parabola open towards the left.
On comparing this equation with y2 = -4ax, we get,
-4a = -8
a = -8/-4 = 2
Thus, co-ordinates 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, then,
x = 2
Length of latus rectum = 4a = 4 (2) = 8
4. x2 = – 16y
Solution: –
Given:
The equation is x2 = -16y
Here we know that the coefficient of y is negative.
So, the parabola opens downwards.
On comparing this equation with x2 = -4ay, we get,
-4a = -16
a = -16/-4
= 4
Thus, co-ordinates of the focus = (0,-a) = (0,-4)
Since, the given equation involves x2, the axis of the parabola is the y-axis.
∴ The equation of directrix, y =a, then,
y = 4
Length of latus rectum = 4a = 4(4) = 16
5. y2 = 10x
Solution: –
Given:
The equation is y2 = 10x
Here we know that the coefficient of x is positive.
So, the parabola open towards the right.
On comparing this equation with y2 = 4ax, we get,
4a = 10
a = 10/4 = 5/2
Thus, co-ordinates of the focus = (a,0) = (5/2, 0)
Since, the given equation involves y2, the axis of the parabola is the x-axis.
∴ The equation of directrix, x =-a, then,
x = – 5/2
Length of latus rectum = 4a = 4(5/2) = 10
6. x2 = – 9y
Solution: –
Given:
The equation is x2 = -9y
Here we know that the coefficient of y is negative.
So, the parabola open downwards.
On comparing this equation with x2 = -4ay, we get,
-4a = -9
a = -9/-4 = 9/4
Thus, co-ordinates of the focus = (0,-a) = (0, -9/4)
Since, the given equation involves x2, the axis of the parabola is the y-axis.
∴ The equation of directrix, y = a, then,
y = 9/4
Length of latus rectum = 4a = 4(9/4) = 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
Solution: –
Given:
Focus (6,0) and directrix x = -6
We know that the focus lies on the x–axis is the axis of the parabola.
So, 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
∴ The equation of the parabola is y2 = 24x.
8. Focus (0,–3); directrix y = 3
Solution: –
Given:
Focus (0, -3) and directrix y = 3
We know that the focus lies on the y–axis, the y-axis is the axis of the parabola.
So, 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
∴ The equation of the parabola is x2 = -12y.
9. Vertex (0, 0); focus (3, 0)
Solution: –
Given:
Vertex (0, 0) and focus (3, 0)
We know that 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.]
The equation of the parabola is of the form y2 = 4ax.
Since, the focus is (3, 0), a = 3
∴ The equation of the parabola is y2 = 4 × 3 × x,
y2 = 12x
10. Vertex (0, 0); focus (–2, 0)
Solution: –
Given:
Vertex (0, 0) and focus (-2, 0)
We know that 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.]
The equation of the parabola is of the form y2=-4ax.
Since, the focus is (-2, 0), a = 2
∴ The equation of the parabola is y2 = -4 × 2 × x,
y2 = -8x
11. Vertex (0, 0) passing through (2, 3) and axis is along x-axis.
Solution: –
We know that the vertex is (0, 0) and the axis of the parabola is the x-axis
The equation of the parabola is either of the from y2 = 4ax or y2 = -4ax.
Given that the parabola passes through point (2, 3), which lies in the first quadrant.
So, the equation of the parabola is of the form y2 = 4ax, while point (2, 3) must satisfy the equation y2 = 4ax.
Then,
32 = 4a(2)
32 = 8a
9 = 8a
a = 9/8
Thus, the equation of the parabola is
y2 = 4 (9/8)x
= 9x/2
2y2 = 9x
∴ The equation of the parabola is 2y2 = 9x
12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Solution: –
We know that the vertex is (0, 0) and the parabola is symmetric about the y-axis.
The equation of the parabola is either of the from x2 = 4ay or x2 = -4ay.
Given that the parabola passes through point (5, 2), which lies in the first quadrant.
So, the equation of the parabola is of the form x2 = 4ay, while point (5, 2) must satisfy the equation x2 = 4ay.
Then,
52 = 4a(2)
25 = 8a
a = 25/8
Thus, the equation of the parabola is
x2 = 4 (25/8)y
x2 = 25y/2
2x2 = 25y
∴ The equation of the parabola is 2x2 = 25y
EXERCISE 11.3
In each of the Exercises 1 to 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.
1. x2/36 + y2/16 = 1
Solution: –
Given:
The equation is x2/36 + y2/16 = 1
Here, the denominator of x2/36 is greater than the denominator of y2/16.
So, 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 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
2. x2/4 + y2/25 = 1
Solution: –
Given:
The equation is x2/4 + y2/25 = 1
Here, the denominator of y2/25 is greater than the denominator of x2/4.
So, 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 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
3. x2/16 + y2/9 = 1
Solution: –
Given:
The 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.
So, 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 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
4. x2/25 + y2/100 = 1
Solution: –
Given:
The equation is x2/25 + y2/100 = 1
Here, the denominator of y2/100 is greater than the denominator of x2/25.
So, 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.
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
5. x2/49 + y2/36 = 1
Solution: –
Given:
The equation is x2/49 + y2/36 = 1
Here, the denominator of x2/49 is greater than the denominator of y2/36.
So, 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 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
6. x2/100 + y2/400 = 1
Solution: –
Given:
The equation is x2/100 + y2/400 = 1
Here, the denominator of y2/400 is greater than the denominator of x2/100.
So, 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.
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
7. 36x2 + 4y2 = 144
Solution: –
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 is along 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
8. 16x2 + y2 = 16
Solution: –
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 is along 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
9. 4x2 + 9y2 = 36
Solution: –
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 is along 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
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)
Solution: –
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
∴ The equation of the ellipse is x2/52 + y2/32 = 1 or x2/25 + y2/9 = 1
11. Vertices (0, ± 13), foci (0, ± 5)
Solution: –
Given:
Vertices (0, ± 13) and foci (0, ± 5)
Here, the vertices are on the y-axis.
So, the equation of the ellipse will be of the form x2/b2 + y2/a2 = 1, where ‘a’ is the semi-major axis.
Then, a =13 and c = 5.
It is known that a2 = b2 + c2.
132 = b2+52
169 = b2 + 15
b2 = 169 – 125
b = √144
= 12
∴ The equation of the ellipse is x2/122 + y2/132 = 1 or x2/144 + y2/169 = 1
12. Vertices (± 6, 0), foci (± 4, 0)
Solution: –
Given:
Vertices (± 6, 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 = 6 and c = 4.
It is known that a2 = b2 + c2.
62 = b2+42
36 = b2 + 16
b2 = 36 – 16
b = √20
∴ The equation of the ellipse is x2/62 + y2/(√20)2 = 1 or x2/36 + y2/20 = 1
13. Ends of major axis (± 3, 0), ends of minor axis (0, ±2)
Solution: –
Given:
Ends of major axis (± 3, 0) and ends of minor axis (0, ±2)
Here, the major axis is along 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 = 3 and b = 2.
∴ The equation for the ellipse x2/32 + y2/22 = 1 or x2/9 + y2/4 = 1
14. Ends of major axis (0, ±√5), ends of minor axis (±1, 0)
Solution: –
Given:
Ends of major axis (0, ±√5) and ends of minor axis (±1, 0)
Here, the major axis is along the y-axis.
So, the equation of the ellipse will be of the form x2/b2 + y2/a2 = 1, where ‘a’ is the semi-major axis.
Then, a = √5 and b = 1.
∴ The equation for the ellipse x2/12 + y2/(√5)2 = 1 or x2/1 + y2/5 = 1
15. Length of major axis 26, foci (±5, 0)
Solution: –
Given:
Length of major axis is 26 and foci (±5, 0)
Since the foci are on the x-axis, the major axis is along 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, 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
∴ The equation of the ellipse is x2/132 + y2/122 = 1 or x2/169 + y2/144 = 1
16. Length of minor axis 16, foci (0, ±6).
Solution: –
Given:
Length of minor axis is 16 and foci (0, ±6).
Since the foci are on the y-axis, the major axis is along the y-axis.
So, the equation of the ellipse will be of the form x2/b2 + y2/a2 = 1, where ‘a’ is the semi-major axis.
Then, 2b =16
b = 8 and c = 6.
It is known that a2 = b2 + c2.
a2 = 82 + 62
= 64 + 36
=100
a = √100
= 10
∴ The equation of the ellipse is x2/82 + y2/102 =1 or x2/64 + y2/100 = 1
17. Foci (±3, 0), a = 4
Solution: –
Given:
Foci (±3, 0) and a = 4
Since the foci are on the x-axis, the major axis is along 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, c = 3 and a = 4.
It is known that a2 = b2 + c2.
a2 = 82 + 62
= 64 + 36
= 100
16 = b2 + 9
b2 = 16 – 9
= 7
∴ The equation of the ellipse is x2/16 + y2/7 = 1
18. b = 3, c = 4, centre at the origin; foci on the x axis.
Solution: –
Given:
b = 3, c = 4, centre at the origin and foci on the x axis.
Since the foci are on the x-axis, the major axis is along 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, b = 3 and c = 4.
It is known that a2 = b2 + c2.
a2 = 32 + 42
= 9 + 16
=25
a = √25
= 5
∴ The equation of the ellipse is x2/52 + y2/32 or x2/25 + y2/9 = 1
19. Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).
Solution: –
Given:
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).
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, where ‘a’ is the semi-major axis.
The ellipse passes through points (3, 2) and (1, 6).
So, by putting the values x = 3 and y = 2, we get,
32/b2 + 22/a2 = 1
9/b2 + 4/a2…. (1)
And by putting the values x = 1 and y = 6, we get,
11/b2 + 62/a2 = 1
1/b2 + 36/a2 = 1 …. (2)
On solving equation (1) and (2), we get
b2 = 10 and a2 = 40.
∴ The equation of the ellipse is x2/10 + y2/40 = 1 or 4x2 + y 2 = 40
20. Major axis on the x-axis and passes through the points (4,3) and (6,2).
Solution: –
Given:
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Since the major axis is on the x-axis, the equation of the ellipse will be the form
x2/a2 + y2/b2 = 1…. (1) [Where ‘a’ is the semi-major axis.]
The ellipse passes through points (4, 3) and (6, 2).
So by putting the values x = 4 and y = 3 in equation (1), we get,
16/a2 + 9/b2 = 1 …. (2)
Putting, x = 6 and y = 2 in equation (1), we get,
36/a2 + 4/b2 = 1 …. (3)
From equation (2)
16/a2 = 1 – 9/b2
1/a2 = (1/16 (1 – 9/b2)) …. (4)
Substituting the value of 1/a2 in equation (3) we get,
36/a2 + 4/b2 = 1
36(1/a2) + 4/b2 = 1
36[1/16 (1 – 9/b2)] + 4/b2 = 1
36/16 (1 – 9/b2) + 4/b2 = 1
9/4 (1 – 9/b2) + 4/b2 = 1
9/4 – 81/4b2 + 4/b2 = 1
-81/4b2 + 4/b2 = 1 – 9/4
(-81+16)/4b2 = (4-9)/4
-65/4b2 = -5/4
-5/4(13/b2) = -5/4
13/b2 = 1
1/b2 = 1/13
b2 = 13
Now substitute the value of b2 in equation (4) we get,
1/a2 = 1/16(1 – 9/b2)
= 1/16(1 – 9/13)
= 1/16((13-9)/13)
= 1/16(4/13)
= 1/52
a2 = 52
Equation of ellipse is x2/a2 + y2/b2 = 1
By substituting the values of a2 and b2 in above equation we get,
x2/52 + y2/13 = 1
EXERCISE 11.4
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.
1. x2/16 – y2/9 = 1
Solution: –
Given:
The equation is x2/16 – y2/9 = 1 or x2/42 – y2/32 = 1
On comparing this equation with the standard equation of hyperbola x2/a2 – y2/b2 = 1,
We get a = 4 and b = 3,
It is known that, a2 + b2 = c2
So,
c2 = 42 + 32
= √25
c = 5
Then,
The coordinates of the foci are (±5, 0).
The coordinates of the vertices are (±4, 0).
Eccentricity, e = c/a = 5/4
Length of latus rectum = 2b2/a = (2 × 32)/4 = (2×9)/4 = 18/4 = 9/2
2. y2/9 – x2/27 = 1
Solution: –
Given:
The equation is y2/9 – x2/27 = 1 or y2/32 – x2/272 = 1
On comparing this equation with the standard equation of hyperbola y2/a2 – x2/b2 = 1,
We get a = 3 and b = √27,
It is known that, a2 + b2 = c2
So,
c2 = 32 + (√27)2
= 9 + 27
c2 = 36
c = √36
= 6
Then,
The coordinates of the foci are (0, 6) and (0, -6).
The coordinates of the vertices are (0, 3) and (0, – 3).
Eccentricity, e = c/a = 6/3 = 2
Length of latus rectum = 2b2/a = (2 × 27)/3 = (54)/3 = 18
3. 9y2 – 4x2 = 36
Solution: –
Given:
The equation is 9y2 – 4x2 = 36 or y2/4 – x2/9 = 1 or y2/22– x2/32 = 1
On comparing this equation with the standard equation of hyperbola y2/a2 – x2/b2 = 1,
We get a = 2 and b = 3,
It is known that, a2 + b2 = c2
So,
c2 = 4 + 9
c2 = 13
c = √13
Then,
The coordinates of the foci are (0, √13) and (0, –√13).
The coordinates of the vertices are (0, 2) and (0, – 2).
Eccentricity, e = c/a = √13/2
Length of latus rectum = 2b2/a = (2 × 32)/2 = (2×9)/2 = 18/2 = 9
4. 16x2 – 9y2 = 576
Solution: –
Given:
The equation is 16x2 – 9y2 = 576
Let us divide the whole equation by 576, we get
16x2/576 – 9y2/576 = 576/576
x2/36 – y2/64 = 1
On comparing this equation with the standard equation of hyperbola x2/a2 – y2/b2 = 1,
We get a = 6 and b = 8,
It is known that, a2 + b2 = c2
So,
c2 = 36 + 64
c2 = √100
c = 10
Then,
The coordinates of the foci are (10, 0) and (-10, 0).
The coordinates of the vertices are (6, 0) and (-6, 0).
Eccentricity, e = c/a = 10/6 = 5/3
Length of latus rectum = 2b2/a = (2 × 82)/6 = (2×64)/6 = 64/3
5. 5y2 – 9x2 = 36
Solution: –
Given:
The equation is 5y2 – 9x2 = 36
Let us divide the whole equation by 36, we get
5y2/36 – 9x2/36 = 36/36
y2/(36/5) – x2/4 = 1
On comparing this equation with the standard equation of hyperbola y2/a2 – x2/b2 = 1,
We get a = 6/√5 and b = 2,
It is known that, a2 + b2 = c2
So,
c2 = 36/5 + 4
c2 = 56/5
c = √(56/5)
= 2√14/√5
Then,
The coordinates of the foci are (0, 2√14/√5) and (0, – 2√14/√5).
The coordinates of the vertices are (0, 6/√5) and (0, -6/√5).
Eccentricity, e = c/a = (2√14/√5) / (6/√5) = √14/3
Length of latus rectum = 2b2/a = (2 × 22)/6/√5 = (2×4)/6/√5 = 4√5/3
6. 49y2 – 16x2 = 784.
Solution: –
Given:
The equation is 49y2 – 16x2 = 784.
Let us divide the whole equation by 784, we get
49y2/784 – 16x2/784 = 784/784
y2/16 – x2/49 = 1
On comparing this equation with the standard equation of hyperbola y2/a2 – x2/b2 = 1,
We get a = 4 and b = 7,
It is know that, a2 + b2 = c2
So,
c2 = 16 + 49
c2 = 65
c = √65
Then,
The coordinates of the foci are (0, √65) and (0, –√65).
The coordinates of the vertices are (0, 4) and (0, -4).
Eccentricity, e = c/a = √65/4
Length of latus rectum = 2b2/a = (2 × 72)/4 = (2×49)/4 = 49/2
In each Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions
7. Vertices (±2, 0), foci (±3, 0)
Solution: –
Given:
Vertices (±2, 0) and foci (±3, 0)
Here, the vertices are on the x-axis.
So, the equation of the hyperbola is of the form x2/a2 – y2/b2 = 1
Since, the vertices are (±2, 0), so, a = 2
Since, the foci are (±3, 0), so, c = 3
It is know that, a2 + b2 = c2
So, 22 + b2 = 32
b2 = 9 – 4 = 5
∴ The equation of the hyperbola is x2/4 – y2/5 = 1
8. Vertices (0, ± 5), foci (0, ± 8)
Solution: –
Given:
Vertices (0, ± 5) and foci (0, ± 8)
Here, the vertices are on the y-axis.
So, the equation of the hyperbola is of the form y2/a2 – x2/b2 = 1
Since, the vertices are (0, ±5), so, a = 5
Since, the foci are (0, ±8), so, c = 8
It is know that, a2 + b2 = c2
So, 52 + b2 = 82
b2 = 64 – 25 = 39
∴ The equation of the hyperbola is y2/25 – x2/39 = 1
9. Vertices (0, ± 3), foci (0, ± 5)
Solution: –
Given:
Vertices (0, ± 3) and foci (0, ± 5)
Here, the vertices are on the y-axis.
So, the equation of the hyperbola is of the form y2/a2 – x2/b2 = 1
Since, the vertices are (0, ±3), so, a = 3
Since, the foci are (0, ±5), so, c = 5
It is known that, a2 + b2 = c2
So, 32 + b2 = 52
b2 = 25 – 9 = 16
∴ The equation of the hyperbola is y2/9 – x2/16 = 1
10. Foci (±5, 0), the transverse axis is of length 8.
Solution: –
Given:
Foci (±5, 0) and the transverse axis is of length 8.
Here, the foci are on x-axis.
The equation of the hyperbola is of the form x2/a2 – y2/b2 = 1
Since, the foci are (±5, 0), so, c = 5
Since, the length of the transverse axis is 8,
2a = 8
a = 8/2
= 4
It is known that, a2 + b2 = c2
42 + b2 = 52
b2 = 25 – 16
= 9
∴ The equation of the hyperbola is x2/16 – y2/9 = 1
11. Foci (0, ±13), the conjugate axis is of length 24.
Solution: –
Given:
Foci (0, ±13) and the conjugate axis is of length 24.
Here, the foci are on y-axis.
The equation of the hyperbola is of the form y2/a2 – x2/b2 = 1
Since, the foci are (0, ±13), so, c = 13
Since, the length of the conjugate axis is 24,
2b = 24
b = 24/2
= 12
It is known that, a2 + b2 = c2
a2 + 122 = 132
a2 = 169 – 144
= 25
∴ The equation of the hyperbola is y2/25 – x2/144 = 1
12. Foci (± 3√5, 0), the latus rectum is of length 8.
Solution: –
Given:
Foci (± 3√5, 0) and the latus rectum is of length 8.
Here, the foci are on x-axis.
The equation of the hyperbola is of the form x2/a2 – y2/b2 = 1
Since, the foci are (± 3√5, 0), so, c = ± 3√5
Length of latus rectum is 8
2b2/a = 8
2b2 = 8a
b2 = 8a/2
= 4a
It is known that, a2 + b2 = c2
a2 + 4a = 45
a2 + 4a – 45 = 0
a2 + 9a – 5a – 45 = 0
(a + 9) (a -5) = 0
a = -9 or 5
Since, a is non – negative, a = 5
So, b2 = 4a
= 4 × 5
= 20
∴ The equation of the hyperbola is x2/25 – y2/20 = 1
13. Foci (± 4, 0), the latus rectum is of length 12
Solution: –
Given:
Foci (± 4, 0) and the latus rectum is of length 12
Here, the foci are on x-axis.
The equation of the hyperbola is of the form x2/a2 – y2/b2 = 1
Since, the foci are (± 4, 0), so, c = 4
Length of latus rectum is 12
2b2/a = 12
2b2 = 12a
b2 = 12a/2
= 6a
It is known that, a2 + b2 = c2
a2 + 6a = 16
a2 + 6a – 16 = 0
a2 + 8a – 2a – 16 = 0
(a + 8) (a – 2) = 0
a = -8 or 2
Since, a is non – negative, a = 2
So, b2 = 6a
= 6 × 2
= 12
∴ The equation of the hyperbola is x2/4 – y2/12 = 1
14. Vertices (±7, 0), e = 4/3
Solution: –
Given:
Vertices (±7, 0) and e = 4/3
Here, the vertices are on the x- axis
The equation of the hyperbola is of the form x2/a2 – y2/b2 = 1
Since, the vertices are (± 7, 0), so, a = 7
It is given that e = 4/3
c/a = 4/3
3c = 4a
Substitute the value of a, we get
3c = 4(7)
c = 28/3
It is known that, a2 + b2 = c2
72 + b2 = (28/3)2
b2 = 784/9 – 49
= (784 – 441)/9
= 343/9
∴ The equation of the hyperbola is x2/49 – 9y2/343 = 1
15. Foci (0, ±√10), passing through (2, 3)
Solution: –
Given:
Foci (0, ±√10) and passing through (2, 3)
Here, the foci are on y-axis.
The equation of the hyperbola is of the form y2/a2 – x2/b2 = 1
Since, the foci are (±√10, 0), so, c = √10
It is known that, a2 + b2 = c2
b2 = 10 – a2 ………….. (1)
It is given that the hyperbola passes through point (2, 3)
So, 9/a2 – 4/b2 = 1 … (2)
From equations (1) and (2), we get,
9/a2 – 4/(10-a2) = 1
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
So, a2 = 5
b2 = 10 – a2
= 10 – 5
= 5
∴ The equation of the hyperbola is y2/5 – x2/5 = 1.
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