JEE Mathematics Parabola MCQs Set 08

Answer: (c) (−1, 2)

Explanation:
1. The axis is the line through the focus perpendicular to the directrix, with direction (1, 1): points (−2 + t, 1 + t).
2. It meets x + y = 3 when −1 + 2t = 3, so t = 2, giving the foot (0, 3).
3. The vertex is the midpoint of the focus and this foot: \( \left(\frac{-2+0}{2},\frac{1+3}{2}\right)=(-1,2) \).

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Answer: (a) 3x + 2y + 14 = 0

Explanation:
1. The foot of the directrix on the axis is the reflection of the focus in the vertex: \( (2\cdot(-1)-2,\ 2\cdot1-3)=(-4,-1) \).
2. The axis has direction (3, 2), so the directrix is perpendicular to it: 3(x + 4) + 2(y + 1) = 0.
3. That is 3x + 2y + 14 = 0.

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Answer: (c) y² = 4x − 8

Explanation:
1. The latus rectum lies on x = 3, so the focus is (3, 0) and the axis is the x-axis.
2. Distance vertex to focus is a = 1, so the parabola is \( y^2=4a(x-2)=4(x-2) \).
3. That is y² = 4x − 8. Check: at x = 3, y² = 4, so y = ±2.

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Answer: (c) pair of lines

Explanation:
1. Group as \( (x+2y)^2-3(x+2y)-4=0 \).
2. Factorise: \( (x+2y-4)(x+2y+1)=0 \).
3. So it is a pair of parallel lines.

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Answer: (b) 4

Explanation:
1. A second-degree equation is a parabola when \( h^2=ab \): \( 2^2=\lambda\cdot1 \), so \( \lambda=4 \).
2. For λ = 4 it is \( (2x+y)^2+2(2x+y)+y+2=0 \), which is not a pair of lines (the discriminant is not zero), so it is a genuine parabola.

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Answer: (a) \( \left(\frac54,1\right) \)

Explanation:
1. Complete the square: \( (y-1)^2=x-1 \).
2. Vertex (1, 1) and 4a = 1, so a = 1/4.
3. Focus \( =(1+a,1)=\left(\frac54,1\right) \).

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Answer: (c) 3y = 2

Explanation:
1. Rewrite as \( 9\left(y-\frac23\right)^2=16x+61 \).
2. The parabola opens along the x-direction, so the axis is the line y = 2/3, i.e. 3y = 2.

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Answer: (b) \( \frac{28}{13} \)

Explanation:
1. Dividing by 169: \( \sqrt{(x-1)^2+(y-3)^2}=\frac{|5x-12y+17|}{13} \), so the focus is (1, 3) and the directrix is 5x − 12y + 17 = 0.
2. Distance from focus to directrix \( =\frac{|5-36+17|}{13}=\frac{14}{13} \).
3. Latus rectum = 2 × this distance = 28/13.

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Answer: (a) x = 0

Explanation:
1. Eliminate t: \( (y-1)^2=4t^2=4(x-1) \).
2. So a = 1 with vertex (1, 1); the directrix is \( x-1=-a \), i.e. x = 0.

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Answer: (a) (32, 32)

Explanation:
1. Here a = 8. The point is \( (at^2,2at)=(8t^2,16t) \) with 16t = −8, so \( t=-\frac12 \).
2. For a focal chord \( t_1t_2=-1 \), so \( t_2=2 \).
3. The other end is \( (8\cdot4,\ 16\cdot2)=(32,32) \).

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Answer: (d) none of these

Explanation:
1. The focus is (2, 0). A line through it meets the parabola in two points unless it is parallel to the axis.
2. The line with slope m = 0 is the axis and meets the parabola at only one point.
3. So the condition is m ≠ 0 (any non-zero real m), which is none of the given ranges.

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Answer: (d) \( b^2c=4a^3 \)

Explanation:
1. Let the chord make angle θ with the axis. Its length is \( c=\frac{4a}{\sin^2\theta} \).
2. Its distance from the vertex is \( b=a\sin\theta \).
3. So \( b^2c=a^2\sin^2\theta\cdot\frac{4a}{\sin^2\theta}=4a^3 \).

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Answer: (a) \( \frac{\sqrt{105}-5}{10} \)

Explanation:
1. Substitute \( x=2y+1 \): \( y^2-2ky-k=0 \), so \( y_1-y_2=2\sqrt{k^2+k} \).
2. The slope of the line gives chord length \( =|y_1-y_2|\sqrt{1+4}=2\sqrt5\sqrt{k^2+k} \).
3. Setting this equal to 4: \( 5(k^2+k)=4 \), i.e. \( 5k^2+5k-4=0 \), so \( k=\frac{-5\pm\sqrt{105}}{10} \).
4. Of the given options only \( \frac{\sqrt{105}-5}{10} \) is a root.

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Answer: (c) c = −4am

Explanation:
1. Make the equation homogeneous using \( 1=\frac{y-mx}{c} \): \( y^2=4ax\cdot\frac{y-mx}{c} \), i.e. \( cy^2-4axy+4amx^2=0 \).
2. The lines through the origin are perpendicular when coefficient of x² plus coefficient of y² is zero: \( 4am+c=0 \).
3. So c = −4am.

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. h
10. o
11. m
12. o
13. g
14. e
15. n
16. i
17. s
18. a
19. t
20. i
21. o
22. n
23.
24. t
25. r
26. i
27. c
28. k
29.
30. f
31. o
32. r
33.
34. t
35. h
36. e
37.
38. p
39. a
40. i
41. r
42.
43. o
44. f
45.
46. l
47. i
48. n
49. e
50. s
51.
52. j
53. o
54. i
55. n
56. i
57. n
58. g
59.
60. t
61. h
62. e
63.
64. o
65. r
66. i
67. g
68. i
69. n
70.
71. t
72. o
73.
74. t
75. h
76. e
77.
78. i
79. n
80. t
81. e
82. r
83. s
84. e
85. c
86. t
87. i
88. o
89. n
90.
91. p
92. o
93. i
94. n
95. t
96. s
97. .
98.
99. P
100. e
101. r
102. p
103. e
104. n
105. d
106. i
107. c
108. u
109. l
110. a
111. r
112.
113. m
114. e
115. a
116. n
117. s
118.
119. t
120. h
121. e
122.
123. s
124. u
125. m
126.
127. o
128. f
129.
130. t
131. h
132. e
133.
134. x
135. ²
136.
137. a
138. n
139. d
140.
141. y
142. ²
143.
144. c
145. o
146. e
147. f
148. f
149. i
150. c
151. i
152. e
153. n
154. t
155. s
156.
157. i
158. s
159.
160. z
161. e
162. r
163. o
164. .

Answer: (a) (0, 2)

Explanation:
1. A ray parallel to the axis passes through the focus after reflection.
2. The vertex is (−1, 2) and a = 1, so the focus is (−1 + 1, 2) = (0, 2).

Teacher's Note:
1. T
2. h
3. e
4.
5. r
6. e
7. f
8. l
9. e
10. c
11. t
12. i
13. n
14. g
15.
16. p
17. r
18. o
19. p
20. e
21. r
22. t
23. y
24.
25. s
26. e
27. n
28. d
29. s
30.
31. r
32. a
33. y
34. s
35.
36. p
37. a
38. r
39. a
40. l
41. l
42. e
43. l
44.
45. t
46. o
47.
48. t
49. h
50. e
51.
52. a
53. x
54. i
55. s
56.
57. t
58. h
59. r
60. o
61. u
62. g
63. h
64.
65. t
66. h
67. e
68.
69. f
70. o
71. c
72. u
73. s
74. .
75.
76. L
77. o
78. c
79. a
80. t
81. e
82.
83. t
84. h
85. e
86.
87. f
88. o
89. c
90. u
91. s
92.
93. f
94. r
95. o
96. m
97.
98. t
99. h
100. e
101.
102. v
103. e
104. r
105. t
106. e
107. x
108.
109. f
110. o
111. r
112. m
113. .

Answer: (d) P is an exterior point while Q is an interior point

Explanation:
1. For P(1, 3): \( y^2-4x=9-4=5>0 \), so P is exterior.
2. For Q(1, 1): \( y^2-4x=1-4=-3<0 \), so Q is interior.

Teacher's Note:
1. S
2. u
3. b
4. s
5. t
6. i
7. t
8. u
9. t
10. e
11.
12. t
13. h
14. e
15.
16. p
17. o
18. i
19. n
20. t
21.
22. i
23. n
24. t
25. o
26.
27. y
28. ²
29.
30. −
31.
32. 4
33. x
34. .
35.
36. N
37. e
38. g
39. a
40. t
41. i
42. v
43. e
44.
45. m
46. e
47. a
48. n
49. s
50.
51. i
52. n
53. s
54. i
55. d
56. e
57.
58. t
59. h
60. e
61.
62. p
63. a
64. r
65. a
66. b
67. o
68. l
69. a
70. ,
71.
72. p
73. o
74. s
75. i
76. t
77. i
78. v
79. e
80.
81. m
82. e
83. a
84. n
85. s
86.
87. o
88. u
89. t
90. s
91. i
92. d
93. e
94. .

Answer: (b) two

Explanation:
1. For (0, −2): \( y^2-4x=4>0 \), so the point is outside the parabola.
2. Two tangents can be drawn from an exterior point.

Teacher's Note:
1. A
2. n
3.
4. e
5. x
6. t
7. e
8. r
9. i
10. o
11. r
12.
13. p
14. o
15. i
16. n
17. t
18.
19. g
20. i
21. v
22. e
23. s
24.
25. t
26. w
27. o
28.
29. t
30. a
31. n
32. g
33. e
34. n
35. t
36. s
37. ,
38.
39. a
40.
41. p
42. o
43. i
44. n
45. t
46.
47. o
48. n
49.
50. t
51. h
52. e
53.
54. c
55. u
56. r
57. v
58. e
59.
60. g
61. i
62. v
63. e
64. s
65.
66. o
67. n
68. e
69. ,
70.
71. a
72. n
73. d
74.
75. a
76. n
77.
78. i
79. n
80. t
81. e
82. r
83. i
84. o
85. r
86.
87. p
88. o
89. i
90. n
91. t
92.
93. g
94. i
95. v
96. e
97. s
98.
99. n
100. o
101. n
102. e
103. .

Answer: (c) y = 3

Explanation:
1. Tangent at (1, 2): \( 2y=2(x+1) \), i.e. y = x + 1.
2. Tangent at (4, 4): \( 4y=2(x+4) \), i.e. \( y=\frac{x+4}{2} \).
3. Equating: 2x + 2 = x + 4, so x = 2 and y = 3. The point (2, 3) lies on y = 3.

Teacher's Note:
1. W
2. r
3. i
4. t
5. e
6.
7. e
8. a
9. c
10. h
11.
12. t
13. a
14. n
15. g
16. e
17. n
18. t
19.
20. w
21. i
22. t
23. h
24.
25. t
26. h
27. e
28.
29. f
30. o
31. r
32. m
33. u
34. l
35. a
36.
37. T
38.
39. =
40.
41. 0
42.
43. a
44. n
45. d
46.
47. s
48. o
49. l
50. v
51. e
52.
53. t
54. h
55. e
56.
57. t
58. w
59. o
60.
61. e
62. q
63. u
64. a
65. t
66. i
67. o
68. n
69. s
70.
71. t
72. o
73. g
74. e
75. t
76. h
77. e
78. r
79. .

Answer: (c) right-angled isosceles

Explanation:
1. The ends of the latus rectum are (a, 2a) and (a, −2a). The tangents are \( y=x+a \) and \( y=-x-a \) with slopes 1 and −1.
2. They are perpendicular and meet on the axis at (−a, 0). Their distances from the point are equal by symmetry.
3. So the triangle is right-angled isosceles, independent of a.

Teacher's Note:
1. T
2. a
3. n
4. g
5. e
6. n
7. t
8. s
9.
10. a
11. t
12.
13. t
14. h
15. e
16.
17. e
18. n
19. d
20. s
21.
22. o
23. f
24.
25. t
26. h
27. e
28.
29. l
30. a
31. t
32. u
33. s
34.
35. r
36. e
37. c
38. t
39. u
40. m
41.
42. h
43. a
44. v
45. e
46.
47. s
48. l
49. o
50. p
51. e
52. s
53.
54. 1
55.
56. a
57. n
58. d
59.
60. −
61. 1
62. .
63.
64. T
65. h
66. e
67. y
68.
69. a
70. r
71. e
72.
73. p
74. e
75. r
76. p
77. e
78. n
79. d
80. i
81. c
82. u
83. l
84. a
85. r
86.
87. a
88. n
89. d
90.
91. s
92. y
93. m
94. m
95. e
96. t
97. r
98. i
99. c
100.
101. a
102. b
103. o
104. u
105. t
106.
107. t
108. h
109. e
110.
111. a
112. x
113. i
114. s
115. .

Answer: (d) π/2

Explanation:
1. A line y = mx meets \( y^2=4x-4 \) where \( y^2-\frac{4}{m}y+4=0 \).
2. Tangency needs zero discriminant: \( \frac{16}{m^2}-16=0 \), so m = ±1.
3. The tangents have slopes 1 and −1, so the angle between them is π/2.

Teacher's Note:
1. S
2. e
3. t
4.
5. t
6. h
7. e
8.
9. d
10. i
11. s
12. c
13. r
14. i
15. m
16. i
17. n
18. a
19. n
20. t
21.
22. o
23. f
24.
25. t
26. h
27. e
28.
29. i
30. n
31. t
32. e
33. r
34. s
35. e
36. c
37. t
38. i
39. o
40. n
41.
42. e
43. q
44. u
45. a
46. t
47. i
48. o
49. n
50.
51. t
52. o
53.
54. z
55. e
56. r
57. o
58. .
59.
60. S
61. l
62. o
63. p
64. e
65. s
66.
67. +
68. 1
69.
70. a
71. n
72. d
73.
74. −
75. 1
76.
77. g
78. i
79. v
80. e
81.
82. p
83. e
84. r
85. p
86. e
87. n
88. d
89. i
90. c
91. u
92. l
93. a
94. r
95.
96. l
97. i
98. n
99. e
100. s
101. .