Question

Class 11 Maths MCQ Questions of Conic Sections with Answers?

Answer 1

Clear all the fundamentals concepts and prepare thoroughly for the exam by taking help from MCQ Questions for class 11 Maths with Answers. To score good marks in the final examination, practice the problems provided here, which will help you to solve the problems in the annual examination.

Class 11 conic sections will incorporate the concept of sections of cone such as

• Parabola
• Ellipse
• Hyperbola
• Circle

Students can Solve the Class 11 Maths MCQ Questions of conic sections with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Conic Sections Objective Questions.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The locus of the point from which the tangent to the circles x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0 are equal is given by the equation

(a) 8x + 19 = 0
(b) 8x – 19 = 0
(c) 4x – 19 = 0
(d) 4x + 19 = 0

2. The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0

(a) 7
(b) 8
(c) 9
(d) 10

3. The center of the ellipse (x + y – 2)² /9 + (x – y)² /16 = 1 is

(a) (0, 0)
(b) (0, 1)
(c) (1, 0)
(d) (1, 1)

4. The parametric coordinate of any point of the parabola y² = 4ax is

(a) (-at², -2at)
(b) (-at², 2at)
(c) (a sin²t, -2a sin t)
(d) (a sin t, -2a sin t)

5. The equation of parabola with vertex at origin the axis is along x-axis and passing through the point (2, 3) is

(a) y² = 9x
(b) y² = 9x/2
(c) y² = 2x
(b) y² = 2x/9

6. At what point of the parabola x² = 9y is the abscissa three times that of ordinate

(a) (1, 1)
(b) (3, 1)
(c) (-3, 1)
(d) (-3, -3)

7. The number of tangents that can be drawn from (1, 2) to x² + y² = 5 is

(a) 0
(b) 1
(c) 2
(d) More than 2

8. In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is

(a) 4/5
(b) 1/3
(c) 3/5
(d) 1/2

9. The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is

(a) 16x² – 9y² – 24xy – 144x + 8y + 224 = 0
(b) 16x² + 9y²– 24xy – 144x + 8y – 224 = 0
(c) 16x² + 9y² – 24xy – 144x – 8y + 224 = 0
(d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

10. The parametric representation (2 + t², 2t + 1) represents

(a) a parabola
(b) a hyperbola
(c) an ellipse
(d) a circle

11. The equation of a hyperbola with foci on the x-axis is

(a) x²/a² + y²/b² = 1
(b) x²/a² – y²/b² = 1
(c) x² + y² = (a² + b²)
(d) x² – y2 = (a² + b²)

12. The equation of parabola with vertex (-2, 1) and focus (-2, 4) is

(a) 10y = x² + 4x + 16
(b) 12y = x² + 4x + 16
(c) 12y = x²  + 4x
(d) 12y = x² + 4x + 8

13. The length of the transverse axis is the distance between the ____. 

(a) Two vertices
(b) Two Foci
(c) Vertex and the origin
(d) Focus and the vertex

14. The length of the latus rectum of x² = -9y is equal to

(a) 3 units
(b) 6 units
(c) 9 units
(d) 12 units

15. The eccentricity of hyperbola is

(a) e =1
(b) e > 1
(c) e < 1
(d) 0 < e < 1

16. The focus of the parabola y² = 8x is

(a) (0,2)
(b) (2,0)
(c) (0,2)
(d) (-2,0)

17. If (a, b) is the mid point of a chord passing through the vertex of the parabola y² = 4x, then

(a) a = 2b
(b) 2a = b
(c) a² = 2b
(d) 2a = b²

18. The line lx + my + n = 0 will touches the parabola y² = 4ax if

(a) ln = am²
(b) ln = am
(c) ln = a² m²
(d) ln = a² m

19. The center of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is? 

(a) (2,-3)
(b) (-2,3)
(c) (-4,6)
(d) (4,-6)

20. The equation of the directrix of the parabola y²+4y+4x+2 = 0 is

(a) x = 1
(b) x = -1
(c) x = 3/2
(d) x = -3/2

Answer 2

1. Answer: (b) 8x – 19 = 0

Explanation: Given equation of circles are x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0

Now, the required line is the radical axis of the two circles are

(x² + y² – 4) – (x² + y² – 8x + 15) = 0

⇒ x²+ y  – 4 – x² – y² + 8x – 15 = 0

⇒ 8x – 19 = 0

2. Answer: (a) 7

Explanation: The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/\(\sqrt{(3^2+4^2)}\)

= {9 + 16 + 10}/\(\sqrt{(9+16)}\)

= 35/\(\sqrt{25}\)

= 35/5

= 7

3. Answer: (d) (1, 1)

Explanation: The center of the given ellipse is the point of intersection of the lines

x + y – 2 = 0 and x – y = 0

After solving, we get

x = 1, y = 1

So, the center of the ellipse is (1, 1)

4. Answer: (c) (a sin²t, -2a sin t) 

Explanation: The point (a sin2t, -2a sin t) satisfies the equation of the parabola y² = 4ax for all values of t. So, the parametric coordinate of any point of the parabola y² = 4ax is (a sin²t, -2a sin t)

5. Answer: (b) y² = 9x/2

Explanation: A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation:

y² = 4ax …………..(i)

Given, point (2,3) lies on the parabola,

⇒ 3² = 4a × 2

⇒ 9 = 4a × 2

⇒ 9/2 = 4a

From equation 1, we get

y² = (9/2)x

⇒ y² = 9x/2

This is the required equation of the parabola.

6. Answer: (b) (3, 1)

Explanation: Given, parabola is x² = 9y

Let P(h, k) is the point on the parabola such that abscissa is 3 times the ordinate.

So, h = 3k ……… 1

Since P(h, k) lies on the parabola

So, h² = 9k ……… 2

From equation 1 and 2, we get

(3k)² = 9k

⇒ 9k² = 9k

⇒ 9k² – 9k = 0

⇒ 9k(k – 1) = 0

⇒ k = 0, 1

When k = 0, h = 0

So k = 1

Now, from equation 1,

h = 3 × 1 = 3

 So, the point is (3, 1)

7. Answer: (b) 1

Explanation: Given point (1, 2) and equation of circle is x² + y² = 5

Now, x²+ y² – 5 = 0

Put (1, 2) in this equation, we get

1² + 2² – 5 = 1 + 4 – 5 

= 5 – 5 

= 0

So, the point (1, 2) lies on the circle.

Hence, only one tangent can be drawn.

8. Answer: (c) 3/5

Explanation: Given, distance between foci = 6

⇒ 2ae = 6

⇒ ae = 3

Again minor axis = 8

⇒ 2b = 8

⇒ b = 4

⇒ b² = 16

⇒ a² (1 – e²) = 16

⇒ a² – a² e² = 16

⇒ a² – (ae)² = 16

⇒ a² – 3² = 16

⇒ a² – 9 = 16

⇒ a² = 9 + 16

⇒ a² = 25

⇒ a = 5

Now, ae = 3

⇒ 5e = 3

⇒ e = 3/5

So, the eccentricity is 3/5

9. Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

Explanation: Given focus S(3, 0) and equation of directrix is: 3x + 4y = 1

⇒ 3x + 4y – 1 = 0

Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix

Then, SP = PM

⇒ SP² = PM²

⇒ (x – 3)² + (y – 0)² = {(3x + 4y – 1 / \(\sqrt{(3^2+4^2})\)

⇒ x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25

⇒ 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x

⇒ 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x

⇒ 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0

 ⇒ 16x2 + 9y² – 24xy – 144x + 8y + 224 = 0

This is the required equation of parabola.

10. Answer: (a) a parabola

Explanation: Let x = 2 + t²

⇒ x – 2 = t² ……….. 1

and y = 2t + 1

⇒ y – 1 = 2t

⇒ (y – 1)/2 = t

From equation 1, we get

x – 2 = {(y – 1)/2}²

⇒ x – 2 = (y – 1)2/4

⇒ (y – 1)² = 4(x – 2)

This represents the equation of a parabola.

Answer 3

11. Answer: (b) x²/a² – y²/b² = 1

Explanation: The equation of a hyperbola with foci on the x-axis is defined as x²/a² – y²/b² = 1 

12. Answer: (b) 12y = x² + 4x + 16 

Explanation: Given, parabola having vertex is (-2, 1) and focus is (-2, 4)

As the vertex and focus share the same abscissa i.e. -2,

parabola axis of symmetry as x = -2

⇒ x + 2 = 0

 Hence, the equation of a parabola is of the type

(y – k) = a(x – h)² where (h, k) is vertex     

Now, focus = (h, k + 1/4a)

Since, vertex is (-2, 1) and parabola passes through vertex

So, focus = (-2, 1 + 1/4a)

Now, 1 + 1/4a = 4

⇒ 1/4a = 4 -1

⇒ 1/4a = 3

⇒ 4a = 1/3

⇒ a = /1(3 × 4)

⇒ a = 1/12

Now, equation of parabola is

(y – 1) = (1/12) × (x + 2)2

⇒ 12(y – 1) = (x + 2)2

⇒ 12y – 12 = x2 + 4x + 4

⇒ 12y = x² + 4x + 4 + 12

⇒ 12y = x² + 4x + 16 

This is the required equation of parabola

13. Answer: (a) Two vertices

Explanation: The length of the transverse axis is the distance between two vertices.

14. Answer: (c) 9 units

Explanation: Given parabola equation: x² = -9y …(i)

Since, the coefficient of y is negative, the parabola opens downwards.

The general equation of parabola is x²= -4ay…(ii)

Comparing (1) and (2), we get

-4a = -9

a = 9/4

We know that the length of latus rectum = 4a = 4(9/4) = 9.

Therefore, the length of the latus rectum of x² = -9y is equal to 9 units.

15. Answer: (b) e > 1

Explanation: The eccentricity of hyperbola is greater than 1. (i.e.) e > 1.

16. Answer: (b) (2,0)

Explanation: Given parabola equation y² = 8x …(i)

Here, the coefficient of x is positive and the standard form of parabola is y² = 4ax …(ii)

Comparing (1) and (2), we get

4a = 8   

a = 8/4 = 2

We know that the focus of parabolic equation y² = 4ax is (a, 0).

Therefore, the focus of the parabola y² =8x is (2, 0).

17. Answer: (d) 2a = b²

Explanation: Let P(x, y) be the coordinate of the other end of the chord OP where O(0, 0)

Now, (x + 0)/2 = a

⇒ x = 2a

and (y + 0)/2 = b

⇒ y = 2b

Now, y² = 4x

⇒ (2b)² = 4 × 2a

⇒ 4b² = 8a

⇒ b² = 2a

18. Answer: (a) ln = am²

Explanation: Given, lx + my + n = 0

⇒ my = -lx – n  

⇒ y = (-l/m)x + (-n/m)

This will touches the parabola y² = 4ax if

(-n/m) = a/(-l/m)

⇒ (-n/m) = (-am/l)

⇒ n/m = am/l

⇒ ln = am²

19. Answer: (a) (2,-3)

Explanation: Given, equation fo the of the circle is 4x² + 4y2 – 8x + 12y – 25 = 0

⇒ x² + y² – 8x/4 + 12y/4 – 25/4 = 0

⇒ x² + y² – 2x + 3y – 25/4 = 0

Now, center = {-(-2), -3} = (2, -3)

20. Answer: (c) x = 3/2

Explanation: Given equation: y² +4y+4x+2 = 0

Rearranging the equation, we get

(y+2)² = -4x+2

(y+2)² = -4(x – (1/2))

Let Y = y+2 and X = x-(1/2)

So, Y² = -4X …(i)

Hence, equation (1) is of the form y² = -4ax. …(ii)

By comparing (1) and (2), we get a=1.

We know that equation of directrix is x= a

Now, substitute a = 1 and x = x-(1/2) in the directrix equation

x – (1/2) = 1

x = 1+(1/2) = 3/2

Therefore, the equation of the directrix of the parabola y²+4y+4x+2 = 0 is 3/2. 

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