Class 11 Maths Chapter 11 Conic Sections MCQs

Students can find MCQs for Class 11 Maths Chapter 11 Conic Sections online, which can assist them in achieving good test results. The multiple-choice questions are based on NCERT guidelines and the CBSE syllabus (2022-2023). The answers and explanations to these multiple-choice questions are comprehensive. Click here to access all chapter-by-chapter MCQs for Class 11 Maths.

MCQs on Class 11 Maths Chapter 11 Conic Sections

Solve the MCQs for Class 11 Maths Chapter 11 Conic Sections. For each MCQ, there are four options, but only one is correct.

Download PDF – Chapter 11 Conic Sections MCQs

Students should choose the best option and compare their results to the ones provided on our website. Also, check the important questions for class 11 Maths as well.

1) The length of the transverse axis is the distance between the ____.

1. Two vertices
2. Two Foci
3. Vertex and the origin
4. Focus and the vertex

Answer: (a) Two vertices

Explanation:

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

2) The parametric equation of the parabola y² = 4ax is

1. x = at; y = 2at
2. x = at²; y = 2at
3. x = at²; y² = at³
4. x = at²; y = 4at

Answer: (b) x = at²; y = 2at

Explanation:

The parametric equation of the parabola y² = 4ax is x = at²; y = 2at.

3) The centre of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is

1. (-2, 3)
2. (1, -3/2)
3. (-4, 6)
4. (4, -6)

Answer: (b) (1, -3/2)

Explanation:

Given circle equation: 4x² + 4y² – 8x + 12y – 25 = 0

⇒ x² + y² – (8x/4) + (12y/4) – (25/4) = 0

⇒ x² +y² -2x +3y -(25/4) = 0 …(1)

As we know that the general equation of a circle is x²+y²+2gx+2fy+c=0, and the centre of the circle = (-g, -f)

Hence, by comparing equation (1) and the general equation,

2g = -2,

Thus, g = -1

2f = 3,  thus, f = 3/2

Now, substitute the values in the centre of the circle (-g, -f), we get,

Centre = (1, -3/2).

Therefore, option (b) (1, -3/2) is the correct answer.

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

1. x = 1
2. x = -1
3. x = 3/2
4. x = -3/2

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 – (½))

Let Y = y+2 and X = x-(½)

So, Y² = -4X …(1)

Hence, equation (1) is of the form y² = -4ax. …(2)

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-(½) in the directrix equation

x – (½) = 1

x = 1+(½) = 3/2.

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

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

1. 0
2. 1
3. 2
4. More than 2

Answer: (b) 1

Explanation:

Given circle equation: x²+y²= 5

x²+y²-5 =0 …(1)

Now, substitute (1, 2) in equation (1), we get

Circle Equation: (1)²+(2)² -5 =0

Equation of circle = 1+5-5 =0

This represents that the point lies on the circumference of a circle, and hence only one tangent can be drawn from (1, 2).

So, option (b) 1 is the correct answer.

6) The length of the latus rectum of x² = -9y is equal to

1. 3 units
2. -3 units
3. 9/4 units
4. 9 units

Answer: (d) 9 units

Explanation:

Given parabola equation: x² = -9y …(1)

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

The general equation of parabola is x²= -4ay…(2)

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.

7) For the ellipse 3x²+4y² = 12, the length of the latus rectum is:

1. 2/5
2. 3/5
3. 3
4. 4

Answer: (c) 3

Explanation:

Given ellipse equation: 3x²+4y² = 12

The given equation can be written as (x²/4) + (y²/3) = 1…(1)

Now, compare the given equation with the standard ellipse equation: (x²/a²) + (y²/b²) = 1, we get

a = 2 and b = √3

Therefore, a > b.

If a>b, then the length of latus rectum is 2b²/a

Substituting the values in the formula, we get

Length of latus rectum = [2(√3)²] /2 = 3

Therefore, the length of the latus rectum of the ellipse 3x²+4y² = 12 is 3.

8) The eccentricity of hyperbola is

1. e =1
2. e > 1
3. e < 1
4. 0 < e < 1

Answer: (b) e > 1

Explanation:

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

9) The focus of the parabola y² = 8x is

1. (0, 2)
2. (2, 0)
3. (0, -2)
4. (-2, 0)

Answer: (b) (2, 0)

Explanation:

Given parabola equation y² = 8x …(1)

Here, the coefficient of x is positive and the standard form of parabola is y² = 4ax …(2)

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).

Hence, option (b) (2, 0) is the correct answer.

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

1. 1/2
2. 1/5
3. 3/5
4. 4/5

Answer: (c) 3/5

Explanation:

Given that the minor axis of ellipse is 8.(i.e) 2b = 8. So, b=4.

Also, the distance between its foci is 6. (i.e) 2ae = 6

Therefore, ae = 6/2 = 3

We know that b² = a²(1-e²)

b² = a² – a²e²

b² = a² – (ae)²

Now, substitute the values to find the value of a.

(4)² = a² -(3)²

16 = a² – 9

a² = 16+9 = 25.

So, a = 5.

The formula to calculate the eccentricity of ellipse is e = √[1-(b²/a²)]

e = √[1-(4²/5²)]

e = √[(25-16)/25]

e = √(9/25) = 3/5.

Hence, option (c) 3/5 is the correct answer.

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