MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers

Conic Sections Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Conic Sections Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 11 Quiz

Class 11 Maths Chapter 11 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Conic Sections Class 11 Maths MCQ online test

Q1.	The straight line y = mx + c cuts the circle x² + y² = a² in real points if
	A.√{a² × (1 + m²)} < c
	B.√{a² × (1 – m²)} < c
	C.√{a² × (1 + m²)} > c
	D.√{a² × (1 – m²)} > c

Ans: √{a² × (1 + m²)} > c
The straight line y = mx + c cuts the circle x² + y² = a² in real points if
√{a² × (1 + m²)} > c

Q2.	Equation of the directrix of the parabola x² = 4ay is
	A.x = -a
	B.x = a
	C.y = -a
	D.y = a

Ans: y = -a
Given, parabola x² = 4ay
Now, its equation of directrix = y = -a

Q3.	The equation of parabola with vertex at origin and directrix x – 2 = 0 is
	A.y² = -4x
	B.y² = 4x
	C.y² = -8x
	D.y² = 8x

Ans: y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

Q4.	The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
	A.7
	B.8
	C.9
	D.10

Ans: 7
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7

Q5.	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)

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

Q6.	If the line 2x – y + λ = 0 is a diameter of the circle x² + y² + 6x − 6y + 5 = 0 then λ =
	A.5
	B.7
	C.9
	D.11

Ans: 9
Given equation of the circle is
x² + y² + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)² + (3)² – 5} = √{9 + 9 – 5} = √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
⇒ -3 × 2 – 3 + λ = 0
⇒ -6 – 3 + λ = 0
⇒ -9 + λ = 0
⇒ λ = 9

Q7.	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

Ans: 1
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.

Q8.	The equation of the circle x² + y² + 2gx + 2fy + c = 0 will represent a real circle if
	A.g² + f² – c < 0
	B.g² + f² – c ≥ 0
	C.always
	D.None of these

Ans: g² + f² – c ≥ 0
Given, equation of the circle is: x² + y² + 2gx + 2fy + c = 0
This equation can be written as
{x – (-g)}² + {y – (-f)}² + = √{g² + f² – c}²
So, the circle is real is g² + f² – c ≥ 0

Q9.	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

Ans: 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
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) /{√(3² + 4²)}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.

Q10.	If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is
	A.2/3
	B.4/3
	C.1/3
	D.4

Ans: 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Q11.	The eccentricity of an ellipse is?
	A.e = 1
	B.e < 1
	C.e > 1
	D.0 < e < 1

Ans: 0 < e < 1
The eccentricity of an ellipse e = √(1 – a²/b²) and 0 < e < 1

Q12.	If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is
	A.(x + 2)² + (y – 3)² = 3²
	B.(x – 2)² + (y + 3)² = 3²
	C.(x – 2)² + (y – 3)² = 3²
	D.(x + 2)² + (y + 3)² = 3²

Ans: (x – 2)² + (y – 3)² = 3²
Radius of the circle = √{(2 – 0)² + (3 – 0)² – 2²}
= √(4 + 9 – 4)
= √9
= 3
So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Q13.	If the length of the major axis of an ellipse is three times the length of the minor axis then its eccentricity is
	A.1/3
	B.1/√3
	C.1/√2
	D.2√2/√3

Ans: 2√2/√3
Given, the length of the major axis of an ellipse is three times the length of the minor axis
⇒ 2a = 3(2b)
⇒ 2a = 6b
⇒ a = 3b
⇒ a² = 9b²
⇒ a² = 9a² (1 – e²) {since b² = a²(1 – e²)}
⇒ 1 = 9(1 – e²)
⇒ 1/9 = 1 – e²
⇒ e² = 1 – 1/9
⇒ e² = 8/9
⇒ e = √(8/9)
⇒ e = 2√2/√3
So, the eccentricity of the ellipse is 2√2/√3

Q14.	The equation of parabola with vertex at origin and directrix x – 2 = 0 is
	A.y² = -4x
	B.y² = 4x
	C.y² = -8x
	D.y² = 8x

Q15.	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/√52
	C.3/5
	D.1/2

Ans: 3/5
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

Q16.	One of the diameters of the circle x² + y² – 12x + 4y + 6 = 0 is given by
	A.x + y = 0
	B.x + 3y = 0
	C.x = y
	D.3x + 2y = 0

Ans: x + 3y = 0
The coordinate of the centre of the circle x² + y² – 12x + 4y + 6 = 0 are (6, -2)
Clearly, the line x + 3y passes through this point.
Hence, x + 3y = 0 is a diameter of the given circle.

Q17.	The center of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is?
	A.(2, -3)
	B.(-2, 3)
	C.(-4, 6)
	D.(4, -6)

Ans: (2, -3)
Given, equation 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)

Q18.	If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is
	A.2/3
	B.4/3
	C.1/3
	D.4

Ans: 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Q19.	The equation of ellipse whose one focus is at (4, 0) and whose eccentricity is 4/5 is
	A.x²/5 + y²/9 = 1
	B.x2 /25 + y² /9 = 1
	C.x²/9 + y²/5 = 1
	D.x²/9 + y²/25 = 1

Ans: x² /25 + y² /9 = 1
Given focus is (4, 0)
⇒ ae = 4
and e = 4/5
a × (4/5) = 4
⇒ a = 5
Now, b² = a² (1 – e²)
⇒ b² = 5² {1 – (4/5)²}
⇒ b² = 25{1 – 16/25}
⇒ b² = 25{(25 – 16)/25}
⇒ b² = 9
Hence, the equation of the ellipse is x²/a² + y²/b² = 1
⇒ x²/5² + y²/9 = 1
⇒ x²/25 + y²/9 = 1