NCERT MCQ CLASS-11 CHAPTER-11 | MATH NCERT MCQ | CONIC SECTIONS | EDUGROWN

In This Post we are  providing Chapter-11 Conic Sections NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON CONIC SECTIONS

Question 1.
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

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

Question 2.
Equation of the directrix of the parabola x² = 4ay is
(a) x = -a
(b) x = a
(c) y = -a
(d) y = a

Answer: (c) y = -a
Given, parabola x² = 4ay
Now, its equation of directrix = y = -a

Question 3.
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 = 8xAnswer

Answer: (c) 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.

Question 4.
The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
(a) 7
(b) 8
(c) 9
(d) 10Answer

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

Question 5.
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)

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

Question 6.
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

Answer: (c) 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

Question 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 2Answer

Answer: (b) 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.

Question 8.
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

Answer: (b) 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

Question 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

Answer: (d) 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.

Question 10.
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

Answer: (b) 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

Question 11.
The eccentricity of an ellipse is?
(a) e = 1
(b) e < 1
(c) e > 1
(d) 0 < e < 1

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

Question 12.
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

Answer: (c) (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

Question 13.
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

Answer: (d) 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

Question 14.
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

Answer: (c) 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.

Question 15.
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

Answer: (c) 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

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