{"id":17486,"date":"2022-05-24T21:30:44","date_gmt":"2022-05-24T16:00:44","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17486"},"modified":"2022-05-16T15:31:18","modified_gmt":"2022-05-16T10:01:18","slug":"mcq-questions-for-class-11-maths-chapter-11","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-11\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Conic Sections Class 11 MCQs Questions 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 Chapter 11 Conic Sections Objective Questions.<\/p>\n

Conic Sections Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Conic Sections Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Conic Sections Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Conic Sections Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nThe straight line y = mx + c cuts the circle x\u00b2 + y\u00b2 = a\u00b2 in real points if
\n(a) \u221a{a\u00b2 \u00d7 (1 + m\u00b2)} < c
\n(b) \u221a{a\u00b2 \u00d7 (1 – m\u00b2)} < c
\n(c) \u221a{a\u00b2 \u00d7 (1 + m\u00b2)} > c
\n(d) \u221a{a\u00b2 \u00d7 (1 – m\u00b2)} > c<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) \u221a{a\u00b2 \u00d7 (1 + m\u00b2)} > c
\nThe straight line y = mx + c cuts the circle x\u00b2 + y\u00b2 = a\u00b2 in real points if
\n\u221a{a\u00b2 \u00d7 (1 + m\u00b2)} > c<\/p>\n<\/details>\n

\n

Question 2.
\nEquation of the directrix of the parabola x\u00b2 = 4ay is
\n(a) x = -a
\n(b) x = a
\n(c) y = -a
\n(d) y = a<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) y = -a
\nGiven, parabola x\u00b2 = 4ay
\nNow, its equation of directrix = y = -a<\/p>\n<\/details>\n

\n

Question 3.
\nThe equation of parabola with vertex at origin and directrix x – 2 = 0 is
\n(a) y\u00b2 = -4x
\n(b) y\u00b2 = 4x
\n(c) y\u00b2 = -8x
\n(d) y\u00b2 = 8x<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) y\u00b2 = -8x
\nSince the line passing through the focus and perpendicular to the directrix is x-axis,
\ntherefore axis of the required parabola is x-axis.
\nLet the coordinate of the focus is S(a, 0).
\nSince 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.
\nSo, 0 = {a – (-2)}\/2
\n\u21d2 0 = (a + 2)\/2
\n\u21d2 a + 2 = 0
\n\u21d2 a = -2
\nThus the coordinate of focus is (-2, 0)
\nLet P(x, y) be a point on the parabola.
\nThen by definition of parabola
\n(x + 2)\u00b2 + (y – 0)\u00b2 = (x – 2)\u00b2
\n\u21d2 x\u00b2 + 4 + 4x + y\u00b2 = x\u00b2 + 4 – 4x
\n\u21d2 4x + y\u00b2 = – 4x
\n\u21d2 y\u00b2 = -4x – 4x
\n\u21d2 y\u00b2 = -8x
\nThis is the required equation of the parabola.<\/p>\n<\/details>\n

\n

Question 4.
\nThe perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
\n(a) 7
\n(b) 8
\n(c) 9
\n(d) 10<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 7
\nThe perpendicular distance = {3 \u00d7 3 – 4 \u00d7 (-4) + 10}\/\u221a(3\u00b2 + 4\u00b2)
\n= {9 + 16 + 10}\/\u221a(9 + 16)
\n= 35\/\u221a25
\n= 35\/5
\n= 7<\/p>\n<\/details>\n

\n

Question 5.
\nThe equation of a hyperbola with foci on the x-axis is
\n(a) x\u00b2\/a\u00b2 + y\u00b2\/b\u00b2 = 1
\n(b) x\u00b2\/a\u00b2 – y\u00b2\/b\u00b2 = 1
\n(c) x\u00b2 + y\u00b2 = (a\u00b2 + b\u00b2)
\n(d) x\u00b2 – y\u00b2 = (a\u00b2 + b\u00b2)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) x\u00b2\/a\u00b2 – y\u00b2\/b\u00b2 = 1
\nThe equation of a hyperbola with foci on the x-axis is defined as
\nx\u00b2\/a\u00b2 – y\u00b2\/b\u00b2 = 1<\/p>\n<\/details>\n

\n

Question 6.
\nIf the line 2x \u2013 y + \u03bb = 0 is a diameter of the circle x\u00b2 + y\u00b2 + 6x \u2212 6y + 5 = 0 then \u03bb =
\n(a) 5
\n(b) 7
\n(c) 9
\n(d) 11<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 9
\nGiven equation of the circle is
\nx\u00b2 + y\u00b2 + 6x \u2212 6y + 5 = 0
\nCenter O = (-3, 3)
\nradius r = \u221a{(-3)\u00b2 + (3)\u00b2 – 5} = \u221a{9 + 9 – 5} = \u221a13
\nSince diameter of the circle passes through the center of the circle.
\nSo (-3, 3) satisfies the equation 2x \u2013 y + \u03bb = 0
\n\u21d2 -3 \u00d7 2 – 3 + \u03bb = 0
\n\u21d2 -6 – 3 + \u03bb = 0
\n\u21d2 -9 + \u03bb = 0
\n\u21d2 \u03bb = 9<\/p>\n<\/details>\n

\n

Question 7.
\nThe number of tangents that can be drawn from (1, 2) to x\u00b2 + y\u00b2 = 5 is
\n(a) 0
\n(b) 1
\n(c) 2
\n(d) More than 2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1
\nGiven point (1, 2) and equation of circle is x\u00b2 + y\u00b2 = 5
\nNow, x\u00b2 + y\u00b2 – 5 = 0
\nPut (1, 2) in this equation, we get
\n1\u00b2 + 2\u00b2 – 5 = 1 + 4 – 5 = 5 – 5 = 0
\nSo, the point (1, 2) lies on the circle.
\nHence, only one tangent can be drawn.<\/p>\n<\/details>\n

\n

Question 8.
\nThe equation of the circle x\u00b2 + y\u00b2 + 2gx + 2fy + c = 0 will represent a real circle if
\n(a) g\u00b2 + f\u00b2 – c < 0
\n(b) g\u00b2 + f\u00b2 – c \u2265 0
\n(c) always
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) g\u00b2 + f\u00b2 – c \u2265 0
\nGiven, equation of the circle is: x\u00b2 + y\u00b2 + 2gx + 2fy + c = 0
\nThis equation can be written as
\n{x – (-g)}\u00b2 + {y – (-f)}\u00b2 + = \u221a{g\u00b2 + f\u00b2 – c}\u00b2
\nSo, the circle is real is g\u00b2 + f\u00b2 – c \u2265 0<\/p>\n<\/details>\n

\n

Question 9.
\nThe equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is
\n(a) 16x\u00b2 – 9y\u00b2 – 24xy – 144x + 8y + 224 = 0
\n(b) 16x\u00b2 + 9y\u00b2 – 24xy – 144x + 8y – 224 = 0
\n(c) 16x\u00b2 + 9y\u00b2 – 24xy – 144x – 8y + 224 = 0
\n(d) 16x\u00b2 + 9y\u00b2 – 24xy – 144x + 8y + 224 = 0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 16x\u00b2 + 9y\u00b2 – 24xy – 144x + 8y + 224 = 0
\nGiven focus S(3, 0)
\nand equation of directrix is: 3x + 4y = 1
\n\u21d2 3x + 4y – 1 = 0
\nLet P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix
\nThen, SP = PM
\n\u21d2 SP\u00b2 = PM\u00b2
\n\u21d2 (x – 3)\u00b2 + (y – 0)\u00b2 = {(3x + 4y – 1) \/{\u221a(3\u00b2 + 4\u00b2)}\u00b2
\n\u21d2 x\u00b2 + 9 – 6x + y\u00b2 = (9x\u00b2 + 16y\u00b2 + 1 + 24xy – 8y – 6x)\/25
\n\u21d2 25(x\u00b2 + 9 – 6x + y\u00b2) = 9x\u00b2 + 16y\u00b2 + 1 + 24xy – 8y – 6x
\n\u21d2 25x\u00b2 + 225 – 150x + 25y\u00b2 = 9x\u00b2 + 16y\u00b2 + 1 + 24xy – 8y – 6x
\n\u21d2 25x\u00b2 + 225 – 150x + 25y\u00b2 – 9x\u00b2 – 16y\u00b2 – 1 – 24xy + 8y + 6x = 0
\n\u21d2 16x\u00b2 + 9y\u00b2 – 24xy – 144x + 8y + 224 = 0
\nThis is the required equation of parabola.<\/p>\n<\/details>\n

\n

Question 10.
\nIf the parabola y\u00b2 = 4ax passes through the point (3, 2), then the length of its latusrectum is
\n(a) 2\/3
\n(b) 4\/3
\n(c) 1\/3
\n(d) 4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 4\/3
\nSince, the parabola y\u00b2 = 4ax passes through the point (3, 2)
\n\u21d2 2\u00b2 = 4a \u00d7 3
\n\u21d2 4 = 12a
\n\u21d2 a = 4\/12
\n\u21d2 a = 1\/3
\nSo, the length of latusrectum = 4a = 4 \u00d7 (1\/3) = 4\/3<\/p>\n<\/details>\n

\n

Question 11.
\nThe eccentricity of an ellipse is?
\n(a) e = 1
\n(b) e < 1
\n(c) e > 1
\n(d) 0 < e < 1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 0 < e < 1
\nThe eccentricity of an ellipse e = \u221a(1 – a\u00b2\/b\u00b2) and 0 < e < 1<\/p>\n<\/details>\n

\n

Question 12.
\nIf the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is
\n(a) (x + 2)\u00b2 + (y – 3)\u00b2 = 3\u00b2
\n(b) (x – 2)\u00b2 + (y + 3)\u00b2 = 3\u00b2
\n(c) (x – 2)\u00b2 + (y – 3)\u00b2 = 3\u00b2
\n(d) (x + 2)\u00b2 + (y + 3)\u00b2 = 3\u00b2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) (x – 2)\u00b2 + (y – 3)\u00b2 = 3\u00b2
\nRadius of the circle = \u221a{(2 – 0)\u00b2 + (3 – 0)\u00b2 – 2\u00b2}
\n= \u221a(4 + 9 – 4)
\n= \u221a9
\n= 3
\nSo, the equation of the circle = (x – 2)\u00b2 + (y – 3)\u00b2 = 3\u00b2<\/p>\n<\/details>\n

\n

Question 13.
\nIf the length of the major axis of an ellipse is three times the length of the minor axis then its eccentricity is
\n(a) 1\/3
\n(b) 1\/\u221a3
\n(c) 1\/\u221a2
\n(d) 2\u221a2\/\u221a3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 2\u221a2\/\u221a3
\nGiven, the length of the major axis of an ellipse is three times the length of the minor axis
\n\u21d2 2a = 3(2b)
\n\u21d2 2a = 6b
\n\u21d2 a = 3b
\n\u21d2 a\u00b2 = 9b\u00b2
\n\u21d2 a\u00b2 = 9a\u00b2 (1 – e\u00b2) {since b\u00b2 = a\u00b2(1 – e\u00b2)}
\n\u21d2 1 = 9(1 – e\u00b2)
\n\u21d2 1\/9 = 1 – e\u00b2
\n\u21d2 e\u00b2 = 1 – 1\/9
\n\u21d2 e\u00b2 = 8\/9
\n\u21d2 e = \u221a(8\/9)
\n\u21d2 e = 2\u221a2\/\u221a3
\nSo, the eccentricity of the ellipse is 2\u221a2\/\u221a3<\/p>\n<\/details>\n

\n

Question 14.
\nThe equation of parabola with vertex at origin and directrix x – 2 = 0 is
\n(a) y\u00b2 = -4x
\n(b) y\u00b2 = 4x
\n(c) y\u00b2 = -8x
\n(d) y\u00b2 = 8x<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) y\u00b2 = -8x
\nSince the line passing through the focus and perpendicular to the directrix is x-axis,
\ntherefore axis of the required parabola is x-axis.
\nLet the coordinate of the focus is S(a, 0).
\nSince 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.
\nSo, 0 = {a – (-2)}\/2
\n\u21d2 0 = (a + 2)\/2
\n\u21d2 a + 2 = 0
\n\u21d2 a = -2
\nThus the coordinate of focus is (-2, 0)
\nLet P(x, y) be a point on the parabola.
\nThen by definition of parabola
\n(x + 2)\u00b2 + (y – 0)\u00b2 = (x – 2)\u00b2
\n\u21d2 x\u00b2 + 4 + 4x + y\u00b2 = x\u00b2 + 4 – 4x
\n\u21d2 4x + y\u00b2 = – 4x
\n\u21d2 y\u00b2 = -4x – 4x
\n\u21d2 y\u00b2 = -8x
\nThis is the required equation of the parabola.<\/p>\n<\/details>\n

\n

Question 15.
\nIn an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is
\n(a) 4\/5
\n(b) 1\/\u221a52
\n(c) 3\/5
\n(d) 1\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 3\/5
\nGiven, distance between foci = 6
\n\u21d2 2ae = 6
\n\u21d2 ae = 3
\nAgain minor axis = 8
\n\u21d2 2b = 8
\n\u21d2 b = 4
\n\u21d2 b\u00b2 = 16
\n\u21d2 a\u00b2 (1 – e\u00b2) = 16
\n\u21d2 a\u00b2 – a\u00b2 e\u00b2 = 16
\n\u21d2 a\u00b2 – (ae)\u00b2 = 16
\n\u21d2 a\u00b2 – 3\u00b2 = 16
\n\u21d2 a\u00b2 – 9 = 16
\n\u21d2 a\u00b2 = 9 + 16
\n\u21d2 a\u00b2 = 25
\n\u21d2 a = 5
\nNow, ae = 3
\n\u21d2 5e = 3
\n\u21d2 e = 3\/5
\nSo, the eccentricity is 3\/5<\/p>\n<\/details>\n

\n

Question 16.
\nOne of the diameters of the circle x\u00b2 + y\u00b2 – 12x + 4y + 6 = 0 is given by
\n(a) x + y = 0
\n(b) x + 3y = 0
\n(c) x = y
\n(d) 3x + 2y = 0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) x + 3y = 0
\nThe coordinate of the centre of the circle x\u00b2 + y\u00b2 – 12x + 4y + 6 = 0 are (6, -2)
\nClearly, the line x + 3y passes through this point.
\nHence, x + 3y = 0 is a diameter of the given circle.<\/p>\n<\/details>\n

\n

Question 17.
\nThe center of the circle 4x\u00b2 + 4y\u00b2 – 8x + 12y – 25 = 0 is?
\n(a) (2, -3)
\n(b) (-2, 3)
\n(c) (-4, 6)
\n(d) (4, -6)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (2, -3)
\nGiven, equation of the circle is 4x\u00b2 + 4y\u00b2 – 8x + 12y – 25 = 0
\n\u21d2 x\u00b2 + y\u00b2 – 8x\/4 + 12y\/4 – 25\/4 = 0
\n\u21d2 x\u00b2 + y\u00b2 – 2x + 3y – 25\/4 = 0
\nNow, center = {-(-2), -3} = (2, -3)<\/p>\n<\/details>\n

\n

Question 18.
\nIf the parabola y\u00b2 = 4ax passes through the point (3, 2), then the length of its latusrectum is
\n(a) 2\/3
\n(b) 4\/3
\n(c) 1\/3
\n(d) 4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 4\/3
\nSince, the parabola y\u00b2 = 4ax passes through the point (3, 2)
\n\u21d2 2\u00b2 = 4a \u00d7 3
\n\u21d2 4 = 12a
\n\u21d2 a = 4\/12
\n\u21d2 a = 1\/3
\nSo, the length of latusrectum = 4a = 4 \u00d7 (1\/3) = 4\/3<\/p>\n<\/details>\n

\n

Question 19.
\nThe equation of ellipse whose one focus is at (4, 0) and whose eccentricity is 4\/5 is
\n(a) x\u00b2\/5 + y\u00b2\/9 = 1
\n(b) x2 \/25 + y\u00b2 \/9 = 1
\n(c) x\u00b2\/9 + y\u00b2\/5 = 1
\n(d) x\u00b2\/9 + y\u00b2\/25 = 1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) x\u00b2 \/25 + y\u00b2 \/9 = 1
\nGiven focus is (4, 0)
\n\u21d2 ae = 4
\nand e = 4\/5
\na \u00d7 (4\/5) = 4
\n\u21d2 a = 5
\nNow, b\u00b2 = a\u00b2 (1 – e\u00b2)
\n\u21d2 b\u00b2 = 5\u00b2 {1 – (4\/5)\u00b2}
\n\u21d2 b\u00b2 = 25{1 – 16\/25}
\n\u21d2 b\u00b2 = 25{(25 – 16)\/25}
\n\u21d2 b\u00b2 = 9
\nHence, the equation of the ellipse is x\u00b2\/a\u00b2 + y\u00b2\/b\u00b2 = 1
\n\u21d2 x\u00b2\/5\u00b2 + y\u00b2\/9 = 1
\n\u21d2 x\u00b2\/25 + y\u00b2\/9 = 1<\/p>\n<\/details>\n

\n

Question 20.
\nThe focus of parabola y\u00b2 = 8x is
\n(a) (2, 0)
\n(b) (-2, 0)
\n(c) (0, 2)
\n(d) (0, -2)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (2, 0)
\nGiven, y\u00b2 = 8x
\nGeneral equation is y\u00b2 = 4ax
\nNow, 4a = 8
\n\u21d2 a = 2
\nNow, focus = (a, 0) = (2, 0)<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download …<\/p>\n

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