Math Assignment Class XI Ch-10 | Conic Sections

ASSIGNMENT ON CONIC SECTIONs CH-10, CLASS-11

Important questions other than NCERT Book, useful for the examination point of view useful for DAV and CBSE board with solution hints.

CIRCLE

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Question 1:   

Question: Find the centre and radius of the circle 3x² + 3y² = 7  

Answer: Centre: (0,0) ; Radius:   

Question 2: 

Find the centre and radius of circle x² + y² – 4x + 10y - 21 = 0

Answer: Centre (2, -5),  Radius =  

Question 3: 

The coordinates of centre of a circle x² + y² + 4x - 6y + 8 = 0

Ans (-2, 3)

Question 4: 

Check whether the following equations represents a circle. If so, determine its centre and radius.

a) 3x² + 3y² + 6x - 4y - 1 = 0

Answer: Centre (-1, 2/3), Radius = 16/9

Here r > 0, yes this equation represents a circle.

b) x² + y² - 12x + 6y + 45 = 0

Answer: Centre = (6, -3), Radius = 0

This equation will represents a point circle because r = 0

c) x² + y² + 4x + 2y + 14 = 0

Answer: Centre = (-2, -1), Radius = -9

Since r < 0, so it will represent an empty set or imaginary circle.

Question 5: 

Find the equation of the circle, the end points of one of whose diameter are (2, -3), (-3, 5).

Answer: Take any arbitrary point P(x, y) on the circle then use diameter form of the eqn. circle we get 

(x - 2)(x + 3) + (y + 3)(y - 5) = 0

x² + y² + x - 2y  = 21

Question 6:

If the lines 2x - 3y = 5 and 3x - 4y =7 are along the diameters of a circle of radius 7 cm, then obtain equation of the circle.

Answer:  Centre(1, -1). Equation of the circle is : (x - 1)² + (y + 1)² = 49

Solution Hint: Point of intersection of the given two lines is the centre of the circle.

Question 7: 

Find the equation of the circle passing through (5, 7), (6, 6), (2, -2). Find its centre and radius.

Answer: Centre = (2, 3), radius = 5 units

Equation of circle is : x² + y² - 4x - 6y  - 12 = 0

Question 8: 

Find the equation of the circle which passes through the points (20, 3), (19, 8) and (2, - 9). Find its centre and radius.

Answer: centre is O(7, 3), Radius = 13, Equation of the circle is (x - 7)² + (y - 3)² = 169

Question 9: 

Find equation of circle whose centre lies on the line x - 4y = 1 and which passes through (3, 7) and (5, 5).

Answer: Equation of circle is : x² + y² + 6x + 2y  - 90 = 0

Question 10: 

One end of the diameter of the circle  x² + y² - 6x + 5y  - 7 = 0 is (-1, 3). Find the other end.

Solution Hint: 

From the given equation find the centre of the circle.

By using mid point formula find the coordinates of other end.

Question 11: 

If line y = √3x + k  touches the circle x² + y² = 16, then find the value of k.

Solution Hint:

Centre of the circle is (0,0). and radius = 4

Now ⊥ distance of the given line from the centre of the circle is = 4

Solve this we get k = ± 8

Question 12: 

Find the equation of the circle whose centre is (3, -1) and which cut off the chord of length 6 units on the line 2x - 5y + 18 = 0. 

Answer: x² + y² -6x + 2y -28  = 0.

Solution Hint: Above problem can be represented in as shown in the figure.

(i) AB = 6 unit, CD is the ⊥ distance of AB from centre C  

(ii) Find CD by using the formula of ⊥ distance we get CD = .

(iii) In △ ACD by using Pythgoras theorem find the radius AC, we get 

(iv) Equation of the circle with radius  and centre (3, -1)  is (x - 3)² + (y + 1)² = 38.

Question 13: 

Find the equation of the circle circumscribed about the triangle whose vertices are (-2, 3), (5, 2), and (6, -1)

Answer Centre (1, -1) radius 5 Equation of circle is x² + y² - 2x + 2y – 23 = 0

Question 14: 

Find the equation of the image of the circle x² + y² + 8x - 16y + 64 = 0 in the line mirror x = 0

Solution Hint

Find the centre of this circle which is = (- 4, 8)  and radius which is = 4

Since y-axis is the mirror so centre of the mirror circle is (4, 8) and radius remain the same.

Required circle is the circle with centre (4, 8) and radius = 4.

So required circle is x² + y² - 8x - 16y + 64 = 0

Question 15: 

Find the equation of the circle which passes through the centre of the circle x² + y² + 8x + 10y - 7 = 0 and is concentric with the circle 2x² + 2y² - 8x - 12y - 9 = 0

Solution Hint:

Find the centre of both the circle which are (-4, -5) and (2, 3)

Now required circle passes through the point A (-4, -5) and its centre is B(2, 3).

Radius of the required circle is |AB| = 10

Required circle is x² + y² - 4x - 6y - 87 = 0

Question 16: 

Find the equation of the circle passing through the point (2, 4) and having centre at the intersection of lines x - 2y = 5 and 3x - y = 5

Solution Hint :

Point of intersection of two lines is O(1, -2)

Circle passing through the point A(2, 4) so radius = |OA| = 

So required eqn of the circle is x² + y² - 2x - 4y - 32 = 0

Question 17: 

Find the equation of the circle of radius 5 units whose centre lies on the y-axis and which passes through the point (3, 2).

Solution Hint: Let centre is O(0, a) and circle passes through the point A(3, 2).

So radius of the circle is |OA| = 5 , ⇒ a = -2 or 6

When a - -2 then eqn. of the circle is x² + y²  - 4y - 21 = 0

When a - -6 then eqn. of the circle is x² + y²  - 12y - 11 = 0

PARABOLA

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Question 1

Find the equation of parabola which is symmetric about the Y-axis and passes through (2, -3).

Answer: 3x² = - 4y 

Question 2

Find the coordinates of a point on the parabola y² = 8x, whose focal distance is 4.

Solution Hint:

From the given parabola find "a" we get a = 2

Focal distance of right handed parabola is |x + a| = |x + 2|

ATQ  |x + 2| = 4 ⇒ x = 2, - 6 (x = - 6 rejected for right handed parabola) so we have x = 2

Putting x = 2 in given eqn. of parabola and find the values of y we get y = ± 4

Required points on the parabola is (2, 4) and (2, - 4)

Question 3

If the line y = mx + 1, is tangent to parabola y² = 4x, find the value of m.

Solution Hint: 

Put the value of y from the eqn. of line to the  eqn. of parabola and find the quadratic equation in terms of x. 

Find Discriminant by using the formula D = b² - 4ac

For tangent  Put D = 0 and find the value of m, we get m = 1

Question 4

Find the equations of the lines joining the vertex of parabola y² = 6x to the point on it having abscissa 24.

Solution Hint

Vertex of the parabola is (0, 0)

Abscissa of the point on the parabola is 24 ⇒ x = 24

Putting x = 24 in y² = 6x we get y = ±12

Required points on the parabola are (24, 12) and (24, -12)

Now eqn. of the line passing through (0, 0) and (24, 12) is x - 2y = 0

Now eqn. of the line passing through (0, 0) and (24, -12) is x + 2y = 0

Question 5

A beam is supported at its ends by supports which are 12 m apart. Since the load is concentrated at its centre, there is a deflection of 3 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm ?.

Solution Hint

Let point on the deflected beam at the deflected position as centre, represented by origin O.

OX is the x-axis and OY is the y-axis

Eqn of parabola is x² = 4ay …. (i)

Point B lies on the parabola so coordinates of point B are (600, 3)

Putting this point in eqn. (i) we get a = 30000

So eqn of parabola becomes x² = 120000y  …… (ii)

Let at point p the deflection is 1 cm. so deflection from the x-axis is 2 cm.

Let distance of point p from y-axis is k m or 100k cm.

So coordinates of point p are (100k, 2).

This point lie on the parabola (ii) so putting this point in eqn (ii) we get k = 2 √6 m.

Question 6

A satellite dish is in the shape of a parabolic surface .Signals from a satellite strike the surface of the dish and are reflected to a single point (the focus) where the receiver is located .The satellite dish shown in the figure has a diameter of 12 meters and a depth of 2 meters. The parabola is positioned in a rectangular coordinate system with its vertex at the origin. How far from the vertex of the dish should the receiver be placed? Also write the equation of the parabolic surface.

Since the parabola is vertical and has its vertex at origin therefore its equation must be of the form x² = 4ay. Point (6, 2) lies on parabola so

36 = 4a(2)

⇒ a = 36/8     ⇒ a = 4. 5

Ans: Receiver should be placed 4.5 m from the vertex.

Equation of parabola x² = 4ay

x² = 4(4. 5)y

x² = 18y   

ELLIPSE

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Question 1

Find the distance between foci for the ellipse x² + 4y² = 1

Answer: Distance between foci is = 2c = √3

Question 2

Find the eccentricity of the ellipse with foci on the x-axis, if its latus rectum be equal to the half of major axis.

Answer:  e = 1/ √2

Question 3

Find the lengths of major and minor axis, the coordinates of the foci, the vertices, the eccentricity and the equations of the directrix of the ellipse 3x² + 2y² = 18.

Answer: Eccentricity (e) = 1/√3, Length of major axis = 6, Length of minor axis = 2√6, Equation of directrix = ± 3√3. Vertices = (0, ±3), Foci = (0, ± √3).

Question 4

Find the coordinates of foci, the vertices, the eccentricity, and length of the latus rectum of the ellipse  

 

Answer: Vertices = (± 7,0), Foci = (± √13, 0), Eccentricity = √13/7, Length of the latus rectum = 72/7

Question 5

Find the equation of the ellipse with major axis along X - axis and passing through (4, 3) and (-1, 4).

Answer: 7x² + 15 y²  = 247  

Question 6

If equation of the ellipse is   9x² + 25 y²  = 225, find its eccentricity.

Answer: e = 4/5

Question 7

Find the equation of ellipse whose eccentricity is 2/3 and latus rectum is 5 and centre is (0,0).

Answer:  

Question 8

Find the equation of an ellipse whose focii are (±4, 0) and eccentricity is 1/3.

Answer:   

Question 9

Find the equation of the ellipse whose centre is at the origin, length of major axis is 9/2 & e = 1/√3  where the major axis is the horizontal axis.

Answer: 16x² + 24y² = 81

Question 10

If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of ellipse.

Answer Length of latus rectum = 39/4

Question 11

Find the equation of the ellipse whose centre is at the origin, foci are (1, 0) and (-1, 0) and eccentricity is ½

Answer:  Required equation of the ellipse is   

HYPERBOLA

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Question 1

Find the equation of hyperbola having the foci (0,  )and passing through the point (2, 3).

Answer   y² - x² = 5

Question 2

Find the eccentricity of hyperbola whose latus rectum is 8, conjugate axis is half of distance between the focii.

Answer: Eccentricity  .

Question 3

Find the equation of the hyperbola whose foci are   and transverse axis of length 10 units.

Answer:    

Question 4

Find the equation of the hyperbola  with foci at (0, ±4) and length of transverse axis  is 6.

Answer: 

Question 5

Find the equation of the hyperbola in standard form with eccentricity √2 and distance between whose foci is 16

Answer: x² – y² = 32

Question 6

Find the equation of the hyperbola having distance between the directrics 4/√3 and passing through the point (2, 1).

Solution Hint:  Distance between the directrix is  2a / e = 4/√3

After complete solution we get two equations of hyperbola 3x² – 2y² = 10 and x² -2y² = 2

Question 7

Find the equation of the set of all points such that the difference of their distances from A (4, 0) and B(- 4, 0) is always equal to 2.

Solution:  Let a point from the set is P(x, y)

A. T. Q,  |AP| - |BP| = 2

Solving this we get equation of hyperbola as 15x² – y² = 15

Question 8

Foci of Hyperbola coincide with the foci of Ellipse     Find the equation of hyperbola. if e = 2

Answer:   

Question 9

If the distance between the foci of  hyperbola is 16 and its eccentricity is √2, then obtain the equation of hyperbola.

Answer: x² -  y²  = 32

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