CBSE Class 11 Mathematics Chapter 11 Conic Sections Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Conic Sections with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Conic Sections MCQs with Answers to know their preparation level.
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MCQ Class 11 Mathematics Conic Sections with Answers - Set - 1
Question 1:
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
Correct Answer – (C)
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 2 :
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
Correct Answer – (B)
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 3 :
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
Correct Answer – (B)
A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation:
y² = 4ax ………….. 1
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.
Question 4 :
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)
Correct Answer – (D)
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)
Question 5 :
The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
(a) 7
(b) 8
(c) 9
(d) 10
Correct Answer – (A)
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7
MCQ Class 11 Mathematics Conic Sections with Answers
Question 6 :
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
Correct Answer – (C)
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
Question 7 :
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)
Correct Answer – (B)
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)
Question 8 :
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)
Correct Answer – (C)
Question 9 :
A man running a race course notes that the sum of the distances from the two flag posts from him is always 10 meter and the distance between the flag posts is 8 meter. The equation of posts traced by the man is
(a) x²/9 + y²/5 = 1
(b) x²/9 + y2 /25 = 1
(c) x²/5 + y²/9 = 1
(d) x²/25 + y²/9 = 1
Correct Answer – (D)
Let the equation of the ellipse be
x²/a² + y²/b² = 1
It is given that the sum of the distances of the man from the two flag posts is 10 m
This means that the sum of focal distances of a point on the ellipse is 10 m
⇒ PS + PS1 = 10
⇒ 2a = 10
⇒ a = 5
Again, given that the distance between the flag posts is 8 meters
⇒ 2ae = 8
⇒ ae = 4
Now, b² = a² (1 – e²)
⇒ b² = a² – a² e²
⇒ b² = a² – (ae)²
⇒ b² = 5² – 4²
⇒ b² = 25 – 16
⇒ b² = 9
⇒ b = 3
Hence, the equation of the path is x²/5² + y²/3² = 1
⇒ x²/25 + y²/9 = 1
Question 10 :
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
Correct Answer – (B)
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 + y² – 4 – x² – y² + 8x – 15 = 0
⇒ 8x – 19 = 0
- NCERT Solutions Class 11 Mathematics Conic Sections with Answers : Exercise 11.1
- NCERT Solutions Class 11 Mathematics Conic Sections with Answers : Exercise 11.2
- NCERT Solutions Class 11 Mathematics Conic Sections with Answers : Exercise 11.3
- NCERT Solutions Class 11 Mathematics Conic Sections with Answers : Exercise 11.4
- NCERT Solutions Class 11 Mathematics Conic Sections with Answers : Exercise 11 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
