NCERT Solutions Class 11 Maths Chapter 10 Conic Sections

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.

 

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.

 

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.

 

Let us consider the equation of the required circle to 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

 

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

 

Let us consider the equation of the required circle to 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

 

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

 

Given:

Centre (-2, 3) and radius 4

Let us consider the equation of a circle with centre (h, k).

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

 

Given:

Centre (1/2, 1/4) and radius 1/12

Let us consider the equation of a circle with centre (h, k).

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

 

Given:

Centre (1, 1) and radius √2

Let us consider the equation of a circle with centre (h, k).

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

 

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

 

The equation of the given 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.

 

Let us consider the equation of the required circle to 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

 

Let us consider the equation of the required circle to 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

 

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.

 

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

 

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.

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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.

 

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 form 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

 

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 form 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

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

 

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

 

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 the major axis = 2a = 2 (5) = 10

Length of the 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

 

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 the major axis = 2a = 2 (4) = 8

Length of the 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

 

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 the major axis = 2a = 2 (10) = 20

Length of the 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

 

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 the major axis = 2a = 2 (7) = 14

Length of the 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

 

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 the major axis = 2a = 2 (20) = 40

Length of the 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

 

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 the major axis = 2a = 2 (6) = 12

Length of the 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

 

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 the major axis = 2a = 2 (4) = 8

Length of the minor axis = 2b = 2 (1) = 2

Eccentricity, e = c/a = √15/4

Length of latus rectum = 2b2/a = (2×12)/4 = 2/4 = ½

 

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 the major axis = 2a = 2 (3) = 6

Length of the 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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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 substituting 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

 

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

Substituting 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

 

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

 

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

 

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

 

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

 

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

 

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

 

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 known 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

 

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 known that, a2 + b2 = c2

So, 22 + b2 = 32

b2 = 9 – 4 = 5

∴ The equation of the hyperbola is x2/4 – y2/5 = 1

 

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 known that, a2 + b2 = c2

So, 52 + b2 = 82

b2 = 64 – 25 = 39

∴ The equation of the hyperbola is y2/25 – x2/39 = 1

 

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

 

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

 

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

 

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

 

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

 

We know that the origin of the coordinate plane is taken at the vertex of the arch, where its vertical axis is along the positive y-axis.

Diagrammatic representation is as follows:

The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).

It is given that at base arch is 10m high and 5m wide.

So, y = 10 and x = 5/2 from the above figure.

It is clear that the parabola passes through point (5/2, 10)

So, x2 = 4ay

(5/2)2 = 4a(10)

4a = 25/(4×10)

a = 5/32

we know the arch is in the form of a parabola whose equation is x2 = 5/8y

We need to find width, when height = 2m.

To find x, when y = 2.

When, y = 2,

x2 = 5/8 (2)

= 5/4

x = √(5/4)

= √5/2

AB = 2 × √5/2m

= √5m

= 2.23m (approx.)

Hence, when the arch is 2m from the vertex of the parabola, its width is approximately 2.23m.

 

Since the height and width of the arc from the centre is 2m and 8m, respectively, it is clear that the length of the major axis is 8m, while the length of the semi-minor axis is 2m.

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, Diagrammatic representation of semi-ellipse is as follows:

The equation of the semi-ellipse will be of the from x2/16 + y2/4 = 1, y ≥ 0 … (1

Let A be a point on the major axis such that AB = 1.5m.

Now draw AC ⊥ OB.

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 get,

(2.5)2/16 + y2/4 = 1

6.25/16 + y2/4 = 1

y2 = 4 (1 – 6.25/16)

= 4 (9.75/16)

= 2.4375

y = 1.56 (approx.)

So, AC = 1.56m

Hence, the height of the arch at a point 1.5m from one end is approximately 1.56m.

 

Let AB be the rod making an angle Ɵ with OX and P(x,y) be the point on it such that

AP = 3cm.

Diagrammatic representation is as follows:

Then, PB = AB – AP = (12 – 3) cm = 9cm [AB = 12cm]

From P, draw PQ ⊥ OY and PR ⊥ OX.

In ΔPBQ, cos θ = PQ/PB = x/9

Sin θ = PR/PA = y/3

we know that, sin2 θ +cos2 θ = 1,

So,

(y/3)2 + (x/9)2 = 1 or

x2/81 + y2/9 = 1

Hence, the equation of the locus of point P on the rod is x2/81 + y2/9 = 1

 

The given parabola is x2 = 12y.

On comparing this equation with x2 = 4ay, we get,

4a = 12

a = 12/4

= 3

The coordinates of foci are S(0,a) = S(0,3).

Now 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

So, the coordinates of A are (-6, 3), while the coordinates of B are (6, 3)

Then, the vertices of ΔOAB are O(0,0), A (-6,3) and B(6,3).

By using the formula,

Area of ΔOAB = ½ [0(3-3) + (-6)(3-0) + 6(0-3)] unit2

= ½ [(-6) (3) + 6 (-3)] unit2

= ½ [-18-18] unit2

= ½ [-36] unit2

= 18 unit2

∴ Area of ΔOAB is 18 unit2

 

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.

So, PA + PB = 10.

We know that if a point moves in plane in such a way that the sum of its distance from two fixed point is constant, then the path is an ellipse, and this constant value is equal to the length of the major axis of the ellipse.

Then, the path described by the man is an ellipse where the length of the major axis is 10m, while points A and B are the foci.

Now let us take the origin of the coordinate plane as the centre of the ellipse, and taking the major axis along the x- axis,

The diagrammatic representation of the ellipse is as follows:

The equation of the ellipse is in the form of x2/a2 + y2/b2 = 1, where ‘a’ is the semi-major axis.

So, 2a = 10

a = 10/2

= 5

Distance between the foci, 2c = 8

c = 8/2

= 4

By using the relation, c = √(a2 – b2), we get,

4 = √(25 – b2)

16 = 25 – b2

b2 = 25 -1

= 9

b = 3

Hence, equation of the path traced by the man is x2/25 + y2/9 = 1

 

Let us consider OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x-axis at point C.

Diagrammatic representation of the ellipse is as follows:

Now let OC = k

From the equation of the given parabola, we have,

So, y2 = 4ak

y = ±2√ak

The 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.

Then,

k2 + (2√ak)2 = (4√ak)2

k2 + 4ak = 16ak

k2 = 12ak

k = 12a

Thus, AB = 4√ak = 4√(a×12a)

= 4√12a2

= 4√(4a×3a)

= 4(2)√3a

= 8√3a

Hence, the side of the equilateral triangle inscribed in parabola y2 = 4ax is 8√3a.

We know that the origin of the coordinate plane is taken at the vertex of the parabolic reflector, where the axis of the reflector is along the positive x – axis.

Diagrammatic representation is as follows:

We know that 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),

y2 = 4ax

102 = 4a(5)

100 = 20a

a = 100/20

= 5

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.

 

We know that 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.

Diagrammatic representation is as follows:

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadways, 18m from the middle.

So, AB = 30m, OC = 6m, and BC = 50m.

The equation of the parabola is of the from 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.

y2 = 4ax

(50)2 = 4a(24)

a = (50×50)/(4×24)

= 625/24

Equation of the parabola, x2 = 4ay = 4×(625/24)y or 6x2 = 625y

The x coordinate of point D is 18.

Hence, at x = 18,

6(18)2 = 625y

y = (6×18×18)/625

= 3.11(approx.)

Thus, DE = 3.11 m

DF = DE +EF = 3.11m +6m = 9.11m

Hence, the length of the supporting wire attached to the roadway 18m from the middle is approximately 9.11m.