# NCERT Solutions for Class 11 Maths

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1.    Given: h = 0, k = 2 and r = 2

Equation of the circle;

2.    Given: h = -2, k = 3 and r = 4

Equation of the circle;

3.    Given:

Equation of the circle;

(x – h)2 + (y – k)2 = r2

4.    Given: h = 1, k = 1 and r = √2

Equation of the circle;

5.    Given: h = -a, k = -b and

Equation of the circle;

6.    Given: Equation of the circle;

(x + 5)2  + (y – 3)2 = 36

⇒ (x + 5)2  + (y – 3)2 = (6)2  …(i)

On comparing eq.(i) with (x – h)2 + (y – k)2 = r2

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

7.    Given: Equation of the circle; x2 + y2 – 4x – 8y – 45 = 0

⇒ (x2 – 4x) + (y2 – 8y) = 45

⇒ (x2 – 4x + 22) + (y2 – 8y + 42) = 45 + 22 + 42

⇒ (x – 2)2 + (y – 4)2 = (√65)2     …(i)

On comparing eq.(i) with (x – h)2 + (y – k)2 = r2

h = 2, k = 4 and r = √65

8.    Given: Equation of the circle;

⇒ (x2 – 8x) + (y2 – 10y) = 12

⇒ (x2 – 8x + 42) + (y2 – 10y + 52) = 12 + 42 + 52

⇒ (x – 4)2 + (y – 5)2 = (√53)2          ….(i)

On comparing eq.(i) with (x – h)2 + (y – k)2 = r2

We get, h = 4, k = 5 and r = √53

9.    Given: Equation of the circle: 2x2 + 2y2 – x = 0

On comparing eq.(i) with (x – h)2 + (y – k)2 = r2

10.   The equation of the circle is (x – h)2 + (y – k)2 = r2     …(i)

∵ Circle passes through point (4, 1)

∴ (4 – h)2 + (1 – k)2 = r2

⇒ 16 + h2 – 8h + 1 + k2 – 2k = r2

⇒ h2 + k2 – 8h – 2k + 17 = r2     ….(ii)

Again Circle passes through point (6, 5)

∴ (6 – h)2 + (5 – k)2 = r2

⇒ 36 + h2 – 12h + 25 + k2 – 10k = r2

⇒ h2 + k2 – 12h – 10k + 61 = r2          ….(iii)

From eq.(ii) and (iii), we have

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

⇒ 4h + 8k = 44

⇒ h + 2k = 11     …(iv)

Since the centre (h, k) of the circle lies on the line 4x + y = 16

∴ 4h + k = 16     …(v)

On solving eq. (iv) and (v), h = 3, k = 4

we have Putting the values of h and k in eq. (ii), we have

32 + 42 – 8 × 3 – 2 × 4 + 17 = r2

⇒ r2 = 9 + 16 – 24 – 8 + 17 = 10

Therefore, the equation of the required circle is

(x – 3)2 + (y – 4)2 = 10

⇒ x2 + 9 – 6x + y + 16 – 8y = 10

⇒ x2 + y2 – 6x – 8y + 15 = 0

11.   The equation of the circle is (x – h)2 + (y – k)2 = r2     …(i)

∵ Circle passes through point (4, 3)

∴ (2 – h)2 + (3 – k)2 = r2

⇒ 4 + h2 – 4h + 9 + k2 – 6k = r2

⇒ h2 + k2 – 4h – 6k + 13 = r2     ….(ii)

Again Circle passes through point (-1, 1)

∴ (-1 – h)2 + (1 – k)2 = r2

⇒ 1 + h2 + 2h + 1 + k2 – 2k = r2

⇒ h2 + k2 + 2h – 2k + 2 = r2          ….(iii)

From eq.(ii) and (iii), we have

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

⇒ -6h – 4k = -11

⇒ 6h + 4k = 11     …(iv)

Since the centre (h, k) of the circle lies on the line x – 3y – 11 = 0

∴ h – 3k = 11     …(v)

On solving eq. (iv) and (v), we have

Putting the values of h and k in eq.(ii), we have

Therefore, the equation of the required circle is

12.  Since the centre of circle lies on axis, therefore the coordinates of centre is (h, 0).

Now the circle passes through the point (2, 3). According to the question,

Taking h = 6, Equation of the circle is (x – 6)2 + (y – 0)2 = (5)2

⇒ x2 + 36 – 12x + y2 = 25

⇒ x2 + y2 – 12x + 11 = 0

Taking h = -2, Equation of the circle is (x + 2)2 + (y – 0)2 = (5)2

⇒ x2 + 4 + 4x + y2 = 25

⇒ x2 + y2 + 4x – 21 = 0.

13.   The circle makes intercepts a with x-axis and b with y-axis.

∴ OA = a and OB = b

∴ Coordinates of A and B are (a, 0) and (0, b) respectively.

Now the circle passes through the points O (0, 0), A(a, 0) and B(0, b).

Putting these coordinates of three points in the equation of the circle,

x2 + y2 + 2gx + 2fy + c = 0     ………(i)

Circle passing through (0,0)

⇒ c = 0

the circle also passes through (a,0) and (0,b)

Putting the values of g, f and c in eq.(i), we have

14.  The equation of the circle is (x – h)2 + (y – k)2 = r2      …(i)

Since the circle passes through point (4, 5) and coordinates of centre are (2, 2).

∴ Radius of circle

Therefore, the equation of the required circle is

15.   Given: Equation of the circle x2 + y2 = 25

⇒ (x – 0)2 + (y – 0)2 = (5)2

On comparing with (x – h)2 + (y – k)2 = r2, we have h = 0, k = 0 and r = 5

Now distance of the point (-2.5, 3.5) from the centre (0, 0)

Therefore, the point (-2.5, 3.5) lies inside the circle.

16.  Given: Equation of parabola

y2 = 12x

Comparing with y2 = 4ax, we have 4a = 12

⇒ a = 3

∴ Coordinates of focus are (3, 0).

Axis of parabola is x-axis or y = 0

Equation of the directrix is x = -3

⇒ x + 3 = 0

Length of latus rectum = 4 × 3 = 12 units

17.  Given: Equation of parabola

x2 = 6y

Comparing with x2 = 4ay, we have 4a = 6

∴ Coordinates of focus are

Axis of parabola is y axis or x = 0

Equation of the directrix is

⇒ 2y + 3 = 0

Length of latus rectum  units

18.  Given: Equation of parabola

y2 = -8x

Comparing with y2 = -4ax we have 4a = 8

⇒ a = 2

∴ Coordinates of focus are (–2, 0).

Axis of parabola is x axis or y = 0

Equation of the directrix is x = 2

⇒ x – 2 = 0

Length of latus rectum = 4 × 2 = 8 units.

19.  Given: Equation of parabola

x2 = -16y

Comparing with x2 = -4ay we have 4a = 16

⇒ a = 4

∴ Coordinates of focus are (0, -4).

Axis of parabola is x axis or x = 0

Equation of the directrix is y = 4

⇒ y – 4 = 0

Length of latus rectum = 4 × 4 = 16 units.

20.  Given: Equation of parabola

y2 = 10x

Comparing with y2 = 4ax, we have 4a = 10

∴ Coordinates of focus are

Axis of parabola is x axis or y = 0

Equation of the directrix is

⇒ 2x + 5 = 0

Length of latus rectum  units.

21.  Given: Equation of parabola

x2 = -9y

Comparing with x2 = -4ay we have 4a = 9

∴ Coordinates of focus are

Axis of parabola is x axis or x = 0

Equation of the directrix is

⇒ 4y – 9 = 0

Length of latus rectum  units.

22.  Given: Directrix: x = -6 and a = 6

Axis of parabola is x-axis

∴ The required equation of parabola y2 = 4ax

⇒ y2 = 4 × 6x

⇒ y2 = 24x

23.  Given: Directrix: y = 3 and a = 3

Axis of parabola is y-axis

∴ The required equation of parabola x2 = -4ay

⇒ x2 = -4 × 3y

⇒ x2 = -12y

24.  Given: vertex is at (0, 0), y = 0 and a = 3

Axis of parabola is x-axis

∴ The required equation of parabola y2 = 4ax

⇒ y2 = 4 × 3x

⇒ y2 = 12x

25.  Given: vertex is at (0, 0), y = 0 and focus is a = 2

Axis of parabola is x-axis

∴ The required equation of parabola

⇒ y2 = -4ax

⇒ y2 = -8x

26.  Given: vertex (0, 0) and axis is along x-axis.

∴ Parabola is of the form y2 = 4ax

Since the parabola passes through the point (2, 3).

∴ The required equation of parabola

27.  Given: vertex (0, 0) and axis is along y-axis.

∴ Parabola is of the form x2 = 4ay

Since the parabola passes through the point (5, 2).

∴ The required equation of parabola

28.  Given: Equation of ellipse:

∵ 36 > 16 ()it’s a horizontal ellipse)

∴ a2 = 36, b2 = 16

⇒ a = 6, b = 4

Now,

∴ Coordinates of foci are (±c, 0)

⇒ (±√5, 0)

Coordinates of vertices are (±a, 0)

⇒ (±6, 0)

Length of major axis = 2a = 2 × 6 = 12 units

Length of major axis = 2b = 2 × 4 = 8 units

Eccentricity

Length of latus rectum   units

29.  Given: Equation of ellipse:

∵ 25 > 4 (it’s a vertical ellipse)

∴ a2 = 25, b2 = 4

⇒ a = 5, b = 2

Now

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±√21)

Coordinates of vertices are (0, ±a)

⇒ (0, ±5)

Length of major axis = 2a = 2 × 5 = 10 units

Length of minor axis = 2b = 2 × 2 = 4 units

Electricity

Length of latus rectum  units.

30.  Given: Equation of ellipse:

∵ 16 > 9 (it’s a horizontal ellipse)

∴ a2 = 16, b2 = 9

⇒ a = 4, b = 3

Now

∴ Coordinates of foci are (±c, 0)

⇒ (±√7, 0)

Coordinates of vertices are (±a, 0)

⇒ (±4, 0)

Length of major axis = 2a = 2 × 4 = 8 units

Length of minor axis = 2b = 2 × 3 = 6 units

Eccentricity

Length of latus rectum  units.

31.  Given: Equation of ellipse:

∵ 100 > 25 (it’s a vertical ellipse)

∴ a2 = 100, b2 = 25

⇒ a = 10, b = 5

Now

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±5√3)

Coordinates of vertices are (0, ±a)

⇒ (0, ±10)

Length of major axis = 2a = 2 × 10 = 20 units

Length of minor axis = 2b = 2 × 5 = 10 units

Eccentricity

Length of latus rectum

32.  Given: Equation of ellipse:

∵ 49 > 36 (it’s a horizontal ellipse)

∴ a2 = 49, b2 = 36

⇒ a = 7, b = 6

Now

∴ Coordinates of foci are (±c, 0)

⇒ (±√13, 0)

Coordinates of vertices are (±a, 0)

⇒ (±7, 0)

Length of major axis = 2a = 2 × 7 = 14 units

Length of minor axis = 2b = 2 × 6 = 12 units

Eccentricity

Length of latus rectum  units.

33.  Given: Equation of ellipse:

∵ 400 > 100 (it’s a vertical ellipse)

∴ a2 = 400, b2 = 100

⇒ a = 20, b = 10

Now

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±10√3)

Coordinates of vertices are (0, ±a)

⇒ (0, ±20)

Length of major axis = 2a = 2 × 20 = 40 units

Length of minor axis = 2b = 2 × 10 = 20 units

Eccentricity

Length of latus rectum   units.

34.  Given: Equation of ellipse: 36x2 + 4y2 = 144

∵ 36 > 4 (it’s a vertical ellipse)

∴ a2 = 36, b2 = 4

⇒ a = 6, b = 2

Now

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±4√2)

Coordinates of vertices are (0, ±a)

⇒ (0, ±6)

Length of major axis = 2a = 2 × 6 = 12 units

Length of minor axis = 2b = 2 × 2 = 4 units

Eccentricity

Length of latus rectum  units.

35.  Given: Equation of ellipse: 16x2 + y2 = 16

∵ 16 > 1 (it’s a vertical ellipse)

∴ a2 = 16, b2 = 1

⇒ a = 4, b = 1

Now

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±√15)

Coordinates of vertices are (0, ±a)

⇒ (0, ±4)

Length of major axis = 2a = 2 × 4 = 8 units

Length of minor axis = 2b = 2 × 1 = 2 units

Eccentricity

Length of latus rectum  units.

36.  Given: Equation of ellipse: 4x2 + 9y2 = 36

∵ 9 > 4 (it’s horizontal ellipse)

∴ a2 = 9, b2 = 4

⇒ a = 3, b = 2

Now

∴ Coordinates of foci are (±c, 0)

⇒ (±√5, 0)

Coordinates of vertices are (±a, 0)

⇒ (±3, 0)

Length of major axis = 2a = 2 × 3 = 6 units

Length of minor axis = 2b = 2 × 2 = 4 units

Eccentricity

Length of latus rectum  units.

37.  Since foci (±4, 0) lie on x-axis, therefore equation of ellipse is

Now Vertices (±a, 0) = (±5, 0)

⇒ a = 5

And Foci (±c, 0) = (±4, 0)

⇒ c = 4

∵ c2 = a2 – b2 ∴ (4)2 = (5)2 – b2

⇒ b2 = 25 – 16 = 9

Therefore, the required equation of ellipse is

38.  Since foci (0, ±5) lie on y-axis, therefore equation of ellipse is

Now Vertices (0, ±a) = (0, ±13)

⇒ a = 13

And Foci (0, ±c) = (0, ±5)

⇒ c = 5

∵ c2 = a2 – b2 ∴ (5)2 = (13)2 – b2

⇒ b2 = 169 – 25 = 144

Therefore, the required equation of ellipse is

39.  Since foci (±4, 0) lie on x-axis, therefore equation of ellipse is

Now Vertices (±a, 0) = (±6, 0)

⇒ a = 6

And Foci (±a, 0) = (±4, 0)

⇒ c = 4

∵ c2 = a2 – b2 ∴ (4)2 = (6)2 – b2

⇒ b2 = 36 – 16 = 20

Therefore, the required equation of ellipse is

40.  Ends of major axis (±3, 0) lie on x-axis, therefore equation of ellipse is

Now Vertices (±a, 0) = (±3, 0)

⇒ a = 3

And Foci (0, ±b) = (0, ±2)

⇒ b = 2

Therefore, the required equation of ellipse is

41.  Ends of major axis (0, ±√5) lie on y-axis, therefore equation of ellipse is

Now Ends of major axis (0, ±a) = (0, ±√5)

⇒ a = √5

And Ends of minor axis (±b, 0) = (±1, 0)

⇒ b = 1

Therefore, the required equation of ellipse is

42.  Since foci (±5, 0) lie on x-axis, therefore equation of ellipse is

∴ Length of major axis = 2a = 26

⇒ a = 13

Foci (±c, 0) = (±5, 0)

⇒ c = 5

c2 = a2 – b2

⇒ (5)2 = (13)2 – b2

⇒ b2 = 169 – 25 = 144

Therefore, the required equation of ellipse is

43.  Since foci (0, ±6) lie on y-axis, therefore equation of ellipse is

∴ Length of major axis = 2b = 16

⇒ b = 8

Foci (0, ±c) = (0, ±6)

⇒ c = 6

c2 = a2 – b2

⇒ (6)2 = a2 – (8)2

⇒ a2 = 36 + 64 = 100

Therefore, the required equation of ellipse is

44.  Since foci (±3, 0) lie on x-axis, therefore equation of ellipse is

∴ Foci (±c, 0) = (±3, 0)

⇒ c = 3

c2 = a2 – b2

⇒ (3)2 = (4)2 – b2

⇒ b2 = 16 – 9 = 7

Therefore, the required equation of ellipse is

45.  Since foci lie on x-axis, therefore equation of ellipse is

∴ c2 = a2 – b2

⇒ (4)2 = a2 – (3)2

⇒ a2 = 16 + 9 = 25

Therefore, the required equation of ellipse is

46.  Since the major axis is along axis, therefore equation of ellipse is

And the ellipse passes through the point (3, 2) therefore     …(i)

And the ellipse passes through the point (1, 6) therefore    …(ii)

Solving eq. (i) and (ii), we have a2 = 40, b2 = 10

Therefore, the required equation of ellipse is

47.  Since the major axis is along axis, therefore equation of ellipse is

And the ellipse passes through the point (4, 3) therefore     ..(i)

And the ellipse passes through the point (6, 2) therefore   …(ii)

Solving eq. (i) and (ii), we have a2 = 52, b2 = 13

Therefore, the required equation of ellipse is

48.  Given: Equation of hyperbola  in the form

Since, the foci and vertices of the hyperbola lies on x-axis.

∴ a2 =16

⇒ a2 = 4, b2 = 9

⇒ b = 3

∴ c2 = a2 + b2

⇒ c2 = 16 + 9 = 25

⇒ c = 25

∴ Coordinates of foci are (±c, 0)

⇒ (±5, 0)

Coordinates of vertices are (±a, 0)

⇒ (±4, 0)

Eccentricity

Length of latus rectum  units.

49.  Given: Equation of hyperbola   in the form

Since, the foci and vertices of the hyperbola lies on y-axis.

∴ a2 = 9

⇒ a = 3, b2 = 27

⇒ b = 3√3

∴ c2 = a2 + b2

⇒ c2 = 9 + 27 = 36

⇒ c = 6

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±6)

Coordinates of vertices are (0, ±a)

⇒ (0, ±3)

Eccentricity

Length of latus rectum  units.

50.  Given: Equation of hyperbola 9y – 4x = 36

Since, the foci and vertices of the hyperbola lies on y-axis.

∴ a2 = 4

⇒ a = 2, b2 = 9

⇒ b = 3

∴ c2 = a2 + b2

⇒ c2 = 4 + 9 = 13

⇒ c = √13

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±√13)

Coordinates of vertices are (0, ±a)

⇒ (0, ±2)

51.  Given: Equation of hyperbola 16x – 9y = 576

Since, the foci and vertices of the hyperbola lies on x-axis.

∴ a2 = 36

⇒ a = 6, b2 = 64

⇒ b = 8

∴ c2 = a2 + b2

⇒ c2 = 36 + 64 = 100

⇒ c = 10

∴ Coordinates of foci are (±c, 0)

⇒ (±10, 0)

Coordinates of vertices are (±a, 0)

⇒ (±6, 0)

52.  Given: Equation of hyperbola 5y2 – 9x2 = 36

Since, the foci and vertices of the hyperbola lies on y-axis.

∴ Coordinates of foci are (0, ±c)

Coordinates of vertices are (0, ±a)

53.  Given: Equation of hyperbola 49y – 16x = 784

Since, the foci and vertices of the hyperbola lies on y-axis.

∴ a2 = 16

⇒ a = 4, b2 = 49

⇒ b = 7

∴ c2 = a2 + b2

⇒ c2 = 16 + 49 = 65

⇒ c = √65

∴ Coordinates of foci are (0, ±c)

⇒ (0, ±√65)

Coordinates of vertices are (0, ±a)

⇒ (0, ±4)

54.  Given: The vertices (±2, 0) lie on x-axis.

∴ The equation of hyperbola in standard form is

Therefore, the required equation of hyperbola is

55.  Given: The vertices (0, ±5) lie on y-axis.

∴ The equation of hyperbola in standard form is

Therefore, the required equation of hyperbola is

56.  Given: The vertices (0, ±3) lie on y-axis.

∴ The equation of hyperbola in standard form is

Therefore, the required equation of hyperbola is

57.  Given: The foci (±5, 0) lie on x-axis.

∴ The equation of hyperbola in standard form is

∴ Length of transverse axis: 2a = 8

Therefore, the required equation of hyperbola is

58.  Given: The foci (0, ±13) lie on y-axis.

∴ The equation of hyperbola in standard form is

∴ Length of conjugate axis 2b = 25

⇒ b = 12

Foci: (c, 0)

= (5, 0) c = 5

Also c2 = a2 + b2

⇒ (13)2 = a2 + (12)2

⇒ a2 = 169 – 144 = 25

Therefore, the required equation of hyperbola is

59.  Given: The foci (±3√5, 0) lie on x-axis.

∴ The equation of hyperbola in standard form is

∴ Foci (±c, 0) = (±3√5, 0)

⇒ c = 3√5

Length of latus rectum:

⇒ b2 = 4a

Also c2 = a2 + b2

⇒ (3√5)2 = a2 + 4a

⇒ a2 + 4a – 45 = 0

⇒ (a + 9) (a – 5) = 0

⇒ a = -9, a = 5 [a = -9 not possible]

⇒ a = 5

⇒ a2 = 25

Length of latus rectm: b2 = 4a

⇒ b2 = 4 × 5 = 20

Therefore, the required equation of hyperbola is

60.  Given: The foci (±4, 0) lie on x-axis.

∴ The equation of hyperbola in standard form is

∴ Foci (±c, 0) = (±4, 0)

⇒ c = 4

Length of latus rectum:

⇒ b2 = 6a

Also c2 = a2 + b2

⇒ (4)2 = a2 + 6a

⇒ a2 + 6a – 16 = 0

⇒ (a + 8) (a – 2) = 0

⇒ a = -8, a = 2 [a = -8 not possible]

⇒ a = 2

⇒ a2 = 4

Length of latus rectm: b2 = 6a

⇒ b2 = 6 × 2 = 12

Therefore, the required equation of hyperbola is

61.  Given: The foci (±7, 0) lie on x-axis.

∴ The equation of hyperbola in standard form is

∴ Vertices (±a, 0) = (±7, 0)

Therefore, the required equation of hyperbola is

62.  Given: The foci (0, ±√10) lie on y-axis.

∴ The equation of hyperbola is standard form is

∴ Foci (0, ±c) = (0, ±√10)

Since given hyperbola passes through (2, 3)

∴ b2 = 10 – a2 = 10 – 18 = -8 [which is not possible]

And b2 = 10 – a2 = 10 – 5 = 5

Therefore, the required equation of hyperbola is

63.  A parabolic reflector with diameter PR = 20 cm and OQ = 5 cm

Vertex of parabola is (0, 0)

Let focus of the parabola be (a, 0)

Now, PR = 20 cm

⇒ PQ = 10 cm

∴ Coordinate of the point P are (5, 10)

Since the point P lies on the parabola y = 4ax2

Hence the focus i.e (5,0) is the mid point of the given diameter.

64.  Let AB be the parabolic arch having O as the vertex and OY as the axis.

The parabola is of the form x2 = 4ay.

Now, CD = 5 m

⇒ OD = 2.5 m

And BD = 10 m

Therefore, coordinates of point B are (2.5, 10).

Since the point B lies on the parabola x2 = 4ay.

∴ Equation of the parabola is

∴ Coordinate of the point Q are

Since the point Q lies on the parabola

Therefore, width of arch is √5 m = 22.4 m approx.

65.  Let AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the roadway.

Now AL = BM = 30 m, OC = 6 m and LM = 100 m

Let O be the vertex and axis of the parabola be y-axis.

Therefore, the equation of the parabola in standard form is x2 = 4ay.

∴ Coordinates of the point B are (50, 24)

Since point B lies on the parabola x2 = 4ay.

Therefore, equation of parabola is

Let length of the supporting wire PQ at a distance of 18 m be h.

∴ OR = 18 m and PR = PQ – QR = PQ – OC = h – 6

∴ Coordinates of point P are (18, h – 6)

Now, since the point P lies on parabola

66.  Given: Width of elliptical arch (AB)

= 2a = 8 ⇒ a = 4 m

Height of the centre (OB) = b = 2 m

The axis of ellipse is x-axis.

Therefore, the equation of ellipse in standard form is

Now, AP = 1.5 m

∴ OP = OA – AP = 4 – 1.5 = 2.5 m

Let PQ = h, then coordinates of Q are (2.5, h)

Since the point Q lies on the ellipse

67.  Let AB be a rod of length 12 cm and (x, y) P be any point on the rod such that PA = 3 cm and PB = 9 cm.

Let AR = a and BQ = b

Then ΔARP ~ ΔPQB

And OB = OQ + BQ = y + b = y + 3y = 4y

In right angles triangle AOB, AB2 = OA 2+ OB2

Which is required locus of point P and which represents an ellipse.

68.  Given: Equation of parabola x2 = 12y which is in the form of x2 = 4ay

∴ 4a = 12

⇒ a = 3

Let AB be the latus rectum of the parabola, then y = 3

∴ x = 4 × 3 × 3 = 36

⇒ x = ±6

The coordinates of A are (-6, 3)

and coordinates of B are (6, 3).

∴ Area of ΔOAB

69.  Let F1 and F2 be two points where the flag parts are fixed on the ground. The origin O is the mid-point of F1F2.

∴ Coordinates of F1 are (-4, 0) and F2 are (4, 0).

Let P (α, β) be any point on the track.

Squaring both sides, we have,

Squaring both sides again, we have,

Which is the required equation of locus of point P.

70.  Given: Equation of the parabola y = 4ax

Let b be the side of an equilateral ΔOAB whose one vertex

is the vertex of parabola and let OC = x

Now, AB = b

Coordinates of point A are

Since, point A lies on the parabola y2 = 4ax

In right angled triangle ΔOAC

Therefore, the side of triangle is 8√3a.

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