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Question-1

Find the equation of the circle with centre (0,2) and radius 2.

Solution:
Here h = 0, k = 2 and y = 2. Therefore, the required equation of the circle is

(x – 0)2 = (y – 2)2 = (2)2

or x2 + y2 – 4y + 4 = 4

or 9; 9;
x2 + y2 – 4y + 4 = 4.

Question-2

Find the equation of the circle with centre (–2,3) and radius 4.

Solution:
Here h = -2, k = 3 and r = 4. Therefore the required equation of the circle is

(x – (-2)2 + (y – 3)2 = (4)2

or (x + 2)2 + (y - 3)2 = 16

or x2 + 4x + 4 + y2 – 6y + 9 = 16 #9.

Question-3

Find the equation of the circle with entre and radius .

Solution:
Here h = = . Therefore the required equation of the circle is

or x2 – x + =

or x2 + y2 – x - = 0

or x2 + y2 – x - = 0

or x2 + y2 – x - = 0

or x2 + y2 – x - = 0

or 36x2 + 36y2 – 36x – 18y + 11 = 0.

Question-4

Find the equation of the circle with centre (1,1) and radius .

Solution:
Here, h = 1, k = 1 and r = . Therefore, the required equation of the circle is

(x - 1)2 + (y -1)2 = ()2

or x2 – 2x + 1 + y2 – 2y + 1 = 2

or x2 + y2 – 2x – 2y = 0.

Question-5

Find the equation of the circle with centre (–a, –b) and radius .

Solution:
Here, h = -a, k = -b and r = . Therefore, the required equation of the circle is

(x – (-a))2 + (y – (-b))2 = (

(x + a)2 + (y + b)2 = a2 – b2

or x2 + 2ax + a2 + y2 + 2by + b2 = a2 – b2

or x2 + y2 + 2ax+ 2by + 2b2 = 0.

Question-6

Find the centre and radius of the circles (x + 5)2 + (y – 3)2 = 36.

Solution:
(x + 5)2 + (y – 3)2 = 36

or (x – (-5))2 + (y - 3)2 = 62

h = -5, k = 3 and r = 6

Therefore, the given circle has centre at (-5, 3) and radius 6.

Question-7

Find the centre and radius of the circles x2 + y2 – 4x – 8y – 45 = 0.

Solution:
The given equation is

x2 + y2 – 4x – 8y – 45 = 0

or (x2 – 4x) + (y2 – 5y) = 45

Now completing the squares with in the parenthesis, we get

(x2 – 4x + 4)+(y2 –8y + 16) = 4 + 16 + 45

or (x - 2)2 + (y – 4)2 = 65

Therefore, the given circle has centre at (2, 4) and radius .

Question-8

Find the centre and radius of the circles x2 + y2 – 8x + 10y – 12 = 0.

Solution:
The given equation is

x2 + y2 – 8x + 10y – 12 = 0

or (x2 – 8x) + (y2 + 10) = 12

or (x2 - 8x + 16) + (y2 + 10y + 25) = 12 + 16+ 25

or (x – 4)2 + (y + 5)2 = 53

Therefore, the given circle has centre at(4, -5) and radius .

Question-9

Find the centre and radius of the circles 2x2 + 2y2 – x = 0.

Solution:
The given equation is

2x2 + 2y2 – x = 0

or x2 + y2 - = 0
or =

or =

or =

Therefore, the given circle has centre at has radius .

Question-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 the equations of the circle be (x-h)2 + (y - k)2 = r2

Since the circle passes through (4, 1) and (6, 5), we have

 

(4 – h)2 + (1 – k)2 = r2

 

and (6 – h)2 + (5 – k)2 = r2

 

Also, since the centre lies on the line 4x + y = 16, we have

 

4h + k = 16

 

Simplifying the equation (i), we get

 

16 – 8h + h2 + 1 – 2k + k2 = r2

 

or 61 – 12h + h2 – 10k + k2 = r2

 

Now,

 

17 – 61 – 8h + 12h + h2 - 2k + 10k + k2 – k2 = 0

 

(eliminating square terms)

 

or -44 + 4h + 8k = 0

 

or 4h + 8k = 44

 

solving equation (iii) and (iv) we get

 

4h + k = 16

 

4h + 8k = 44

 

- - -

 

-7k = -28 k = 4

 

Substituting k = 4 in equation (iii) we get

 

4h + 4 = 16

 

or 9; 9; 9; 4h = 12

 

or h = 3

 

Substituting the value of h = 3, k = 4 in equation

 

(i) we get

 

(4 – 3)2 + (1 – 4)2 = r2

 

or 1 + 9 = r2

 

or 10 = r2

 

Hence, the required equation of the circle is

 

(4 – 3)2 + (y – 4)2 = 10

 

or x2 – 6x + 9 +y2 – 8y + 16 = 10

 

or x2 – y2 – 6x – 8y + 15 = 0.

Question-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 the equation of the circle be (x – h)2 + (y – k)2 = r2

Since the circle passes through (2, 3) and (-1, 1) we have

(2 – h)2 + (3 – k)2 = r2 ……(i)

and (-1 – h)2 + (1 – k)2 = r2 ……(ii)

Also since the centre lies on the line x – 3y – 11 = 0, we have

h – 3k = 11 …….(iii)

Simplifying the equation (i) and (ii) we have

(i) 4 – 4h + h2 + 9 – 6k + k2 = r2

13 – 4h + h2 – 6k + k2 = r2

(ii) 1 + 2h + h2 + 1 – 2k + k2 = r2

2 + 2h + h2 – 2k + k2 = r2

Eliminating square terms, we get

13 – 4h + h2 – 6k + k2 = 2 + 2h + h2 – 2k + k2

or 13 – 2 – 4h – 2h + h2 – h2 – 6k + 2k + k2 – k2 = 0

or 11 – 6h – 4k = 0

or 6h + 4k = 11 ….(iv)

Solving (iii) and (iv) we get

6h – 18k = 66

6h + 4k = 11

- - -

– 22k = 55

or k = =

putting the value of k in (iii), we have

h – = 11

or h + = 11

or h = 11 – = =

Putting the value of (h, k) in (i), we get

+ = r2

or = r2

or = r2

or = r2

or = r2

or r2 =

or r2 =

Hence, the required equation of the circle is

=

or x2 – 7x +

or x2 – 7x + y2 + 5y =

or x2 + y2 – 7x + 5y = 14

or x2 + y2 – 7x + 5y – 14 = 0.

Question-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 the equation of the circle be (x – h)2 + (y – k)2 = r2

Since the circle passes through, (2, 3) and its radius is 5

(2 – h)2 + (3 – k)2 = 25

or 4 – 4h + h2 + 9 – 6k + k2 = 25

or – 12 - 4h + h2 – 6k + k2 = 0 …..(i)

Also since the centre lies on the x-axis, we have

K = 0 …..(ii)

Putting k = 0 in equation (i) we have

-12 – 4h + h2 = 0

or h2 – 4h – 12 = 0

or h2 – 6h + 12 – 12 = 0

or h(h – 6) + 2(h – 6) = 0

or (h – 6)(h + 2) = 0

or h = 6 or h = -2

Hence, the required equations of the circle are

a) For h = 6

(x – 6)2 + (y – 0)2 = (5)2

or (x2 - 12x + 36 + y2 = 25

or x2 + 4x + 4 + y2 = 25

or x2 + y2 + 4x – 21 = 0.

Question-13

Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Solution:

Let the equation of the circle be

x2 + y2 + 2gx + 2fy + c = 0

Since it passes through the origin (0,0)

Substituting (0, 0) in this equation

 

c = 0

Substituting (0, a) in this equation

 

a2 + 2af = 0

2af = -a2

f = = -

Substituting (b, 0) in this equation

b2 + 2gb = 0

2gb = -b2

g = = -

Substituting the value of f and g in the general equation

x2 + y2 + n + = 0

Therefore the equation of the required circle is

x2 + y2 – bx – ay = 0

Question-14

Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Solution:
Since the centre is (2, 2)

We have (x – 2)2 + (y – 2)2 = r2

This circle passes through the point (4, 5)

(4 – 2)2 + (5 – 2)2 = r2

or 4 + 9 = r2

or r2 = 13

The required equation of the circle is

(x – 2)2 + (y – 2)2 = 13

or x2 – 4x + 4 + y2 – 4y + 4 = 13

or ; x2 + y2 – 4x – 4y - 5 = 0.

Question-15

Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Solution:
The centre of the circle x2 + y2 = 25 is (0, 0).

Let us write it as x2 + y2 – 25 = 0 and take f(x, y) = x2 + y2 – 25. For the points lying on the circle, f(x, y) = 0, for the points inside circle, it has sign as the sign obtained by putting the value of the coordinates of the centre in the expression. If it has opposite sign then it lies outside circle.

F(0, 0) = 0 + 0 – 25 = -ve.

Given point is (-2.5, 3.5)

Hence f(-2.5, 3.5) = (-2.5)2 + (3.5)2 – 25

                         = 6.25 + 12.5 – 25

                         = 18.5 – 25

                         = -6.5 = -ve

Hence, the given point lies inside the circle.

Question-16

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. y2 = 12x

Solution:
The given equation involves y2, so the axis of symmetry is along the x-axis.

The coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y2 = 4ax, we find that a = 3. Thus, the focus of the parabola is (3, 0) and the equation of the directrix of the parabola is x = - 3. Length of the Latus rectum LL' is 4a = 4 × 3 = 12.

Question-17

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. x2 = 6y

Solution:
The given equation involves x2, so the axis symmetry is along y-axis.

The coefficient of y is positive so that parabola opens upward. Comparing with the given equation x2 = 4ay, we find that a = Thus, the focus of the parabola is and the equation of the directrix of the parabola is y =

Length of the latus rectum LL' is 4a = 4 × = 6.

Question-18

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. y2 = – 8x

Solution:
The given equation involves y2, so the axis of symmetry is along the x-axis.

The coefficient of x is negative, so the parabola opens to the left. Comparing with the given equation y2 = 4ax, we find that a = 2. Thus the focus of the parabola is (- 2, 0) and the equation of the directrix of the parabola is x = 2. Length of the Latus rectum LL' is 4a = 4 × 2 = 8.

 

Question-19

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. x2 = – 16y

Solution:
The given equation involves x2, so the axis of symmetry is along the y-axis.

The coefficient of y is negative, so the parabola opens downward. Comparing with the given equation x2 = - 4ay, we find that a = 4. Thus the focus of the parabola is (0, - 4) and the equation of the directrix of the parabola is y = 4. Length of the Latus rectum LL' is 4a = 4 × 4 = 16.

Question-20

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. y2 = 10x

Solution:
The given equation involves y2, so the axis of symmetry is x-axis

y2 =

The coefficient of x is positive, so the parabola opens to the right.

 

Focus = , equation of directrix is x = and length of latus rectum = 4a =
4 × = 10, equation of the axis is y = 0.

Question-21

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. x2 = -9y

Solution:
The given equation is x2 = -9y

The given equation involves x2, so the axis of symmetry is y-axis.

x2 = -4 -4ay

where a = > 0

The coefficient of y is negative. The parabola opens downwards.

Focus = , equation of directrix y = , the length of latus rectum

        = 4a = 4 × = 9 and the equation of axis is x = 0.

Question-22

Find the equation of the parabola that satisfies the given conditions: Focus (6,0); directrix x = – 6

Solution:
Since the focus (6, 0) lies on the x-axis, the x-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form of either y2 = 4ax or y2 = - 4ax. Since the directrix is x = -6 and the focus is (6, 0) the parabola is to be form of y2 = 4ax with
a = 6. Hence, the required equation is y2 = 4(6)x = 24x.

Question-23

Find the equation of the parabola that satisfies the given conditions: Focus (0,–3); directrix y = 3

Solution:
Since the focus (0, - 3) lies on the y-axis, the y-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form, either x2 = 4ay or x2 = - 4ay. As the directrix is y = 3, the equation of the parabola is x2 = -4 × 3ay = -12y or x2 = -12y.

Question-24

Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0); focus (3, 0)

Solution:
Since the vertex is at (0,0) and the focus is at (3,0) which lies on x-axis, the x-axis is the axis of the parabola. Therefore, equation of the parabola of the form

y2 = 4ax y2 = 4(3)x= 12x.

Question-25

Find the equation of the parabola that satisfies the given conditions: Vertex (0,0); focus (–2,0)

Solution:
Since the vertex is at (0,0) and the focus is at (- 2, 0) which lies on x-axis, the x-axis is the axis of the parabola. Therefore, equation of the parabola is of the form

y2 = - 4ax y2 = - 4(2)x = - 8x

Question-26

Find the equation of the parabola that satisfies the given conditions: Vertex (0,0) passing through (2,3) and axis is along x-axis.

Solution:
Since the parabola is symmetric about x-axis and has its vertex at the origin, the equation is of the form y2 = 4ax or y2 = -4ax, where the sign depends on whether the parabola opens to the right or left. But the parabola passes through (2, 3) which lies in the first quadrant, it must open to the right. Thus the equation is of the form y2 = 4ax.

Since the parabola passes through (2, 3) we have

32 = 4a(2) 9 = 8a a =

Therefore, the equation of the parabola is

y2 = 4 = x 2y2 = 9x.

Question-27

Find the equation of the parabola that satisfies the given conditions: Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis.

Solution:
Since the parabola is symmetric about y-axis and has its vertex at the origin, the equation is of the form x2 = 4ay or x2 = -4ay, where the sign depends on whether the parabola opens upwards or downwards. But the parabola passes through (5, 2) which lies in the first quadrant, it must open upwards. Thus, the equations is of the form x2 = 4ay.

Since the parabola passes through (5, 2), we have

52 = 4a(2)

25 = 8a

a =

Therefore, the equation of the parabola is

x2 = 4 = 2x2 = 25y.

Question-28

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

Solution:
Since denominator of x2 is larger than the donominator of y2, the major axis is along the x-axis. Comparing the given equation with = 1.

We get a = 6 and b = 4

Also = =

Therefore, the coordinates of foci, (-c, 0) and (c, 0) are and , Vertices, (-a, 0) and (-6, 0) and (6, 0). Length of the major axis, 2a is 12 units.

The length of the mirror axis, 2b is 8 units and eccentricity, e = is = .

The length of latus rectum = = units.

Question-29

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

Solution:
Since the denominator of y2 is larger than the denominator of x2, the major axis is along the y-axis. Comparing the given equation with the standard equation

= 1

we have b = 2 and a = 5

Also c = = =

And e = =

Hence, the foci, (0, c) and (0, c) are (0, ) and (0, -; the vertices, (0, a) and (0,-a) are(0, 5) and (0, -5); length of the major axis, 2a is 10 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is The length of latus rectum = =

Question-30

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

Solution:
The given equation is = 1 ……(i)

Here, denominator of x2 > denominator of y2.

So, the foci of the ellipse are on the x-axis. Let a2 = 16 and b2 = 9

We get a = 4 and b = 3

Also, c = = =

Therefore, the coordinates of the foci, (-c, 0) and (c, 0) are (-,0) and (,0); the vertices, (-a, 0) and (a, 0) are (-4, 0). Length of the major axis, 2a is 8 units. The length of the minors axis, 2b is 6 units and eccentricity, e = is . The length of latus rectum = =

Question-31

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

Solution:
The given equation is = 1

Here, denominator of x2 < denominator of y2.

The foci of the ellipse are on the y-axis. Comparing the given equation with the standard equation

= 1

we have b = 5 and a = 10.

Also c = = = = 5

And c = = =

Hence, the foci (0, c) and (0, c) are (0, 5) and (0, 5), the vertices (0, a) and (0, -a) are (0, 10) and (0, -10), length of the major axis 2a is 20 units, the length of the minor axis 2b is 10 units and the eccentricity of the ellipse is The length of the latus rectum = = 5 units.

Question-32

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

Solution:
The given equation is = 1

Here, denominator of x2 > denominator of y2

The foci of the ellipse are on the x-axis.

Comparing the given equation with the standard equation

= 1

We have a = 7 and b = 6

Also c = = =

c = =

Hence, the foci,(c, 0) and (-c, 0) and (,0) and (-,0); the vertices, (a, 0) and (-a, 0) are (7, 0) and (-7, 0); the length of the major axis, 2a is 14 units, the length of the minor axis, 2b is 12 units and eccentricity of the ellipse, e is .
The length of latus rectum is = units.

Question-33

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

Solution:
The given equation is = 1

Here, denominator of x2 < denominator of y2

The foci of the of ellipse are on the y –axis.

Let b2 = 100 and a2 = 400

b =10 and a = 20

Also c = = = = 10 and e = .

Hence, the foci (0, c) and (0, -c) are (0, 10 and (0, -10, the vertices (0, a) and

(0, -a) are (0, 20) and (0, -20), the length of the major axis 2a is 40 units, the length of the minor axis 2b is 20 units and eccentricity of the ellipse is The length of latus rectum is = 10 units.

Question-34

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
36x2 + 4y2 = 144.

Solution:
The given equation of the ellipse can be written in standard form as.

= 1 or = 1

since the denominator of x2 < denominator of y2, the major axis is along the y-axis. Comparing the given equation with the standard equation

= 1

we have b = 2 and a = 6

Also, c = = = = 4

and e = = = =

Hence the foci, (0, c) and (0, c) are (0, 4) and (0, -4). Vertices, (0, a) and (0, a) are (0,6) and (0,-6); the length of the major axis, 2a is 12 units; the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is .

The length of latus rectum, is = units.

Question-35

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
16x2 + y2 = 16.

Solution:
The given equation of the ellipse can be written in standard form as

or = 1

Since the denominator of x2 < denominator of y2, the major axis is along the y-axis. Comparing the given equation with the standard equation.

= 1

we have b = 1 and a = 4

Also, e = =

Hence the foci, (0, c) and (0, -c) are(0, and (0, -; vertices, (0, a) and (0, a) are (0, 4) and (0, -4); the length of the major axis, 2b is 2 units and the eccentricity of the ellipse, e is .

The length of latus rectum = = = .

Question-36

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
4x2 + 9y2 = 36.

Solution:
The given equation of the ellipse can be written in standard from as

Since the denominator of x2 > denominator of y2, the major axis is along the x-axis. Comparing the given equation with the standard equation

we have a = 3 and b = 2

Also, c =

And e = =

Hence the foci, (c, 0) and (-c, 0) are ( and (-; vertices (a, 0) and (-a, 0) are (3, 0) and (-3, 0); the length of the major axis, 2a is 6 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is .

The length of latus rectum = = units.

Question-37

Find the equation for the ellipse that satisfies the given conditions: Vertices (± 4, 0). Foci (± 4, 0)

Solution:
Since the vertices are on x-axis the equation will be of the form

Where a is the semi-major axis.

Given that a = 5, c = ± 4

Therefore, from the relation

c2 = a2 – b2, we get

16 = 25 – b2

or b = 3

Hence, the equation of the ellipse is

= 1.

Question-38

Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ± 13) foci (0, ± 5)

Solution:
Since the vertices are on y-axis the equation will be of the form

where a is the semi-major axis.

Given that a = 13, c = ± 5

Therefore, from the relation

c2 = a2 – b2, we get

25 = 169 – b2

or b2 = 144 or b = 12

Hence, the equation of the ellipse to

 

= 1.

Question-39

Find the equation for the ellipse that satisfies the given conditions: Vertices (± 6, 0) foci (± 4, 0)

Solution:
Since the vertices are on x-axis, the equation will be of the form

= 1

where a is the semi-major axis.

Given that a = 6, c = ± 4

Therefore from the relation

c2 = a2 – b2, we get

16 = 36 – b2

b2 = 20 or b = 2

Hence, the equation of the ellipse is

= 1.

Question-40

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis(± 3, 0), ends of minor axis (0, ± 2).

Solution:
Since the vertices are on x-axis, the equation will be of the form

 

= 1

Where a is the semi – major axis.

Given a = 3, b = 2

Hence, the equation of the ellipse is

= 1.

Question-41

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ± ), ends of minor axis (± 1, 0)

Solution:
Since the vertices are on y-axis, the equation will be of the form

= 1

Given a = and b = 1

Hence the equation of the ellipse is

= 1.

Question-42

Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci( ± , 0)

Solution:
Since the foci are on the x-axis, the major axis will be on x-axis. The equation will be of the form

= 1

Given, 2a = 26 or a = 13 and c = ± 5

We know that c2 = a2 – b2 or 25 – 169 – b2

or

Hence, the equation of the ellipse is

= 1.

Question-43

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci(0, +6)

Solution:
Since the foci are on the y-axis, the major axis will be on y-axis. The equation will be of the form

= 1

Given 2b = 16 or b = 8 and c = 16

We know that c2 = a2 – b2 or 36 – a2 – 64

or a2 = 100 or a = 10

Hence, the equation of the ellipse is

= 1.

Question-44

Find the equation for the ellipse that satisfies the given conditions: Foci
(± 3, 0), a = 4

Solution:
Since the foci are on the x-axis, the major axis will be on x-axis. The equation will be of the form

= 1

Given c = ± 3; a = 4

Therefore, c2 = a2 – b2 9 – 16 – b

or b2 = 7 b =

Hence, the equation of then ellipse is

= 1.

Question-45

Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on a x axis.

Solution:
Since foci are on the x-axis, the equation will be of the form

= 1

Given that b = 3, c = 4

Therefore c2 = a2 – b2

or 16 = a2 – 9 a2 = 25

or a = 5

Hence, the equation of the ellipse is

= 1.

Question-46

Find the equation for the ellipse that satisfies the given conditions: Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Solution:
Since the major axis is on the y-axis, the equation will be of the form

= 1

Given that the points(3, 2) and (1, 6) lie on the ellipse, we have

= 1 and = 1

or 9a2 + 4b2 = a2b2

and a2 + 36b2 = a2b2

or 9a2 + 4b2 = a2b2

and 9a2 + 324b2 = 9a2b2

On subtracting we get

ca2 + 4b2 = a2b2

9a2 + 324b2 = 9a2b2

- - ; ; -

- 320b2 = -8a2b2

or a2 = 40

Hence, the required equation is of form

= 1

To find b2 we substitute a2 = 40 in = 1

or = 1

or

or

or b2 = 10

Hence the equation of the given ellipse is

= 1

Question-47

Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4,3) and (6,2).

Solution:
Since the major axis is on the x-axis, the equation will be of the form

= 1 and = 1

or 16b2 + 9a2 = a2b2

and 36b2 + 4a2 = a2b2

or 64b2 + 36a2 = 4a2b2

and 324b2 + 36a2 = 9a2b2

On subtracting we get

64b2 + 36a2 = 4a2b2

324b2 + 36a2 = 9a2b2

- - -

260b2 = -5a2b2

 

To find b2 we substitute a2 – 52 in = 1

or ; = 1

or ; or b2 = 13

Hence, the required equation is = 1.

Question-48

= 1

Solution:
Comparing the equation = 1 with the standard equation = 1.

We have, a = 4, b = 3

and c = = = = 5

Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 4, 0). Also, the eccentricity e = = and the length of latus rectum

= = = units.

Question-49

= 1

Solution:
Comparing the equation = 1 with the standard equation = 1

We have, a = 3, b = 3

and c = = = = 6

 

Therefore, the coordinates of the foci are (0, ± 6) and that of vertices are (0, ± 3). Also, the eccentricity e = = and the length of the latus rectum = = = = 18 units.

Question-50

9y2 – 4x2 = 36.

Solution:
Dividing the equation by 26 on both sides, we have

= 1 = 1

Comparing the equation with the standard equation = 1, we find that a = 2, b = 3 and c = = =

Therefore, the coordinates of the foci are (0, ± ) and that of the vertices are (0, ± 2). Also the eccentricity e = = and the length of the latus rectum = = = 9 units.

Question-51

16x2 – 9y2 = 576.

Solution:
Dividing the equation by 576 on both sides, we have

= 1 or = 1

Comparing the equation with the standard equation

= 1

we find that a = 6, b = 8 and

c = = = = 10

Therefore, the coordinates of foci are (± 10, 0) and that of vertices are (± 6, 0). Also, the eccentricity e = = and the length of latus rectum = = = units.

Question-52

5y2 – 9y2 = 36.

Solution:
Dividing the equation by 36 on both sides, we have

= 36 = 1

Comparing the equation with the standard equation = 1, we find that a = ,
b = 2 and c = = = Therefore the coordinates of foci are and that of vertices are . Also the eccentricity e = = = and the length of latus rectum is = = units.

Question-53

49y2 – 16x2 = 784.

Solution:
Dividing the equation by 784 on both sides, we have

= 1 = 1

Comparing the equation with the standard equation = 1, we find t hat d = 4, b = 7 and c = = =

Therefore the coordinates of foci are (0, ± ) and that of vertices are (0, ± 4). Also the eccentricity e = =and the length of latus rectum is

=

Question-54

Vertices (± 2, 0), foci (± 3, 0)

Solution:
Since the foci are on x-axis, the equation of the hyperbola is of the form

= 1

Given : vertices are (± 2, 0), a = 2

Also, since foci are (± 3, 0), c = 3 and b2 = c2 – a2 = 9 – 4 = 5

Therefore, the equation of the hyperbola is

= 1.

Question-55

Vertices (0, ± 5), foci (0, ± 8)

Solution:
Since the foci are on y axis, the equations of the hyperbola is of the form

= 1

Given: vertices are (0, ± 5), a = 5

Also, since foci are (0, ± 8), c = 8 and b2 = c2 – a2 = 64 – 25 = 39

Therefore, the equation of the hyperbola is = 1.

Question-56

Vertices (0, ± 3), foci (0, ± 5)

Solution:
Since the foci are on y-axis the equation of the hyperbola is of the form

= 1

Given: vertices are (0, ± 3), a = 3

Also, since foci are(0, ± 5), c = 5 and b2 = c2 – a2 = 25 – 9 = 16

Therefore, the equation of the hyperbola is

= 1.

Question-57

Vertices (± 5, 0), the transverse axis is of length.

Solution:
Since the foci are on x-axis, the equation of the hyperbola is of the form

= 1

Given: foci are (± 5, 0), c = 5

Length of transverse axis = 2a = 8 a = 4

Therefore b2 = c2 – a2 = 25 – 16 = 9

= 1

9x2 – 16y2 = 144.

Question-58

Foci (0, ± 13), the conjugate is of length 24.

Solution:
Since the foci are on y-axis, the equation of the hyperbola is of the form

= 1

Given: foci are (0, ± 13), c = 13

Length of conjugate axis = 2b = 24 b = 12

or b2 = c2 – a2 or 144 169 – a2

or a2 = 25 or a = 5

The equation of the hyperbola is

= 1

or 144y2 - 25x2 = 3600.

Question-59

Foci (± 3,0), the latus rectum is of length.

Solution:
Since the foci are on x-axis, the equation of the hyperbola is of the form
= 1

Given: foci are (± 3,0), c = 3

and length of latus rectum = = 8

As b2 = 4a

We have c2 = a2 + b2

or 45 = a2 + 4a

or a2 + 4a – 45 = 0

or a2 + 9a – 5a - 45 = 0

or a(a + 9) –5(a – 9) = 0

or (a + 9) (a – 5) = 0

or a = -9 or a = 5)

since a cannot be negative, we take a = 5 and so b2 = 20.

Therefore, the equation of the required hyperbola is

= 1 4x2 – 5y2 = 100.

Question-60

Foci (± 4, 0), the latus rectum is of length 12.

Solution:
Since the fonci are on x-axis, the equation of the hyperbola is of the form

= 1

Given: foci are (± 4, 0), c = 4

and length of latus rectum = = 12

or 3x2 – y2 = 12

or c2 – a2 + b2

or 16 = a2 + 6a

or a2 + 6a – 16 = 0

or a2 + 8a – 2a – 16 = 0

or a(a + 8) – 2(a + 8) = 0

or (a + 8) (a - 2) = 0

or a = -8 or a = 2

since a cannot be negative, we take a = 2 and so b2 = 12

Therefore the equation of the required hyperbola is

 

= 1 3x2 – y2 = 12.

Question-61

Vertices(± 7, 0), e =

Solution:
Here a = 7 and e =

But e = = or c =

Also, c2 = a2 + b2 = 49 + b2

or = b2

or = b2

or = b2 or = b2

or b2 =

Therefore, the equation of the required hyperbola

= 1 or = 1

or = 1 or 7x2 – 9y2 = 343.

Question-62

Foci (0, ± ), passing through (2, 3)

Solution:
Since the foci are on y-axis, the equation the hyperbola is of the form

= 1
Since foci are (0, ± ), c =

Given that point(2, 3) lies on the hyperbola, we have

= 1 9b2 – 4a2 = a2b2
Also, c2 = a2 + b2 or 9b2 – 4a2 = a2b2

or b2 = 10 – a2

substituting the value of b2 in

9b2 – 4a2 = a2b2 we get

or 9(10 – a2) – 4a2 = a2(10 – a)2

or 90 – 9a2 - 4a2 = 10a2 – a4

or 90 – 23a2 – a4 = 0

or 90 – 18a2 – 5a2 + a4 = 0

or 18(5 – a2)(18 – a2) = 0

From a2 + b2 = 10 we have

5 + b2 = 10 we have

5 + b2 = 10 or 18 + b2 = 10

or b2 = 5 or b2 = -8

Rejecting –ve value, we have b2 = 5 and a2 = 5. Therefore, the equation of the required hyperbola is

 

= 1 y2 – x2 = 5.


 




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