Category: Class 11 Mathematics Objective Questions

250+ TOP MCQs on Conic Sections – Parabola-1 & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Conic Sections – Parabola-1”.

1. Find the focus of parabola with equation y2=100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: c
Clarification: Comparing equation with y2=4ax.
4a=100 => a=25.
Focus is at (a, 0) i.e. (25, 0).

2. Find the focus of parabola with equation y2=-100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: d
Clarification: Comparing equation with y2=-4ax.
4a=100 => a=25.
Focus is at (-a, 0) i.e. (-25, 0).

3. Find the focus of parabola with equation x2=100y.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: a
Clarification: Comparing equation with x2=4ay.
4a=100 => a=25.
Focus is at (0, a) i.e. (0, 25).

4. Find the focus of parabola with equation x2=-100y.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: b
Clarification: Comparing equation with x2=-4ay.
4a=100 => a=25.
Focus is at (0, -a) i.e. (0, -25).

5. Find the equation of latus rectum of parabola y2=100x.
a) x=25
b) x=-25
c) y=25
d) y=-25
Answer: a
Clarification: Comparing equation with y2=4ax.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is x=a => x=25.

6. Find the equation of latus rectum of parabola y2=-100x.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: b
Clarification: Comparing equation with y2=-4ax.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is x=-a => x=-25.

7. Find the equation of latus rectum of parabola x2=100y.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: d
Clarification: Comparing equation with x2=4ay.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is y=a => y=25.

8. Find the equation of latus rectum of parabola x2=-100y.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: c
Clarification: Comparing equation with x2=-4ay.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is y=-a => y=-25.

9. Find the equation of directrix of parabola y2=100x.
a) x=25
b) x=-25
c) y=25
d) y=-25
Answer: b
Clarification: Comparing equation with y2=4ax.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is x=-a => x=-25.

10. Find the equation of directrix of parabola y2=-100x.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: a
Clarification: Comparing equation with y2=-4ax.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is x=a => x=25.

250+ TOP MCQs on Statistics – Mean Deviation & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Statistics – Mean Deviation”.

1. Find the mean deviation about the median of the scores of a batsman given below.

Innings Scores
1 20
2 56
3 0
4 84
5 11
6 120

a) 10
b)10.5
c) 11
d) 9
Answer: b
Clarification: Mean deviation = (frac{1}{n})[Σi = n |xI – A|], where A is median or AM
From the given data, Median, A = (20 + 56)/2 = 38
⇒ Mean deviation = 1/6 x (63) = 10.5.

2. What is the mean deviation from the mean for the following data?

a) 0
b) 3
c) 1
d) ½
Answer: a
Clarification: Mean = (117 + 156 + 206 + 198 + 223)/5 = 180

Xi 117 156 206 198 223
Xi – mean -63 -24 26 18 43

Mean deviation = (frac{1}{n})[Σi = 5 |xI – mean|] = 1/5 x [(-63) + (-24) + 26 + 18 + 43] = 1/5 x [0] = 0.

3. The mean deviation of an ungrouped data is 150. If each observation is increased by 3.5%, then what is the new mean deviation?
a) 153.5
b) 3.5
c) 155.25
d) 150
Answer: c
Clarification: If x1, x2, …, xn are the observations, then the new observations are (1.035) x1, (1.035) x2, ……, (1.035) xn.
Therefore, the new mean is (1.035) x̄
Now, Mean deviation = (frac{1}{n})[Σi = n |xI – mean|]
⇒ New mean deviation = (frac{1}{n})[Σi = n|(1.035)xI – (1.035) x̄|] = (1.035) × (frac{1}{n})[Σi = n |xI – mean|] = 1.035 x 150 = 155.25.

4. Find the mean deviation about mean from the following data:

xi 3 5 20 25 27
fi 5 12 20 8 15

a) 7.7
b) 15
c) 8.7
d) 6.2
Answer: a
Clarification: From the given data,

xi fi fixi |xi-18| fi|xi-18|
3 5 15 15 75
5 12 60 13 156
20 20 400 2 40
25 8 200 7 56
27 15 405 9 135
Σ fi = 60 Σ fixi = 1080 Σ fi|xi – 15| = 462

Now, Mean = (frac{1}{n}) Σ fixi = 1080/60 = 18
⇒ Mean deviation = (frac{1}{n}) Σ fi|xi – 18| = 462/60 = 7.7.

5. What is the geometric mean of 5,52, ….,5n?
a) 5n/2
b) 5(n+1)/2
c) 5n(n+1)/2
d) 5n
Answer: b
Clarification: Geometric Mean = (5 x 52 x …… x 5n)1/n = [5(1+2+…+n)]1/n = [5n(n+1)/2]1/n = 5(n+1)/2.

6. In a class there are 20 juniors, 15 seniors and 5 graduate students. If the junior averaged 65 in the midterm exam, the senior averaged 70 and the graduate students averaged 91, then what is the mean of the centre class approximately?
a) 71
b) 74
c) 70
d) 72
Answer: c
Clarification: Combined mean = (Σ xini)/(Σ ni) = (20 × 65 + 15 × 70 + 5 × 91)/(20 + 15 + 5) = 70.

7. Find the mean deviation from mean of the observations: a, a+d, …., (a+2nd).
a) n(n + 1)d2/3
b) n(n + 1)d2/2
c) a + n(n + 1)d2/2
d) n(n + 1)d/(2n + 1)
Answer: d
Clarification: Mean = (frac{1}{n}) Σ xi = (frac{1}{2n+1}) [a + (a + d) + … + (a + 2nd)] = a + nd
⇒ Mean Deviation = (frac{1}{2n+1}) [2 × d × (1 + 2 + … + n)] = [n (n + 1) (d)]/(2n + 1).

250+ TOP MCQs on The Empty Set & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “The Empty Set”.

1. Find the odd one out.
a) Null set
b) Void set
c) Infinite set
d) Empty set
Answer: c
Clarification: Null set, void set, or empty set is a set which contains no elements. Infinite set is a set having infinite number of elements.

2. Which of the following is representation of empty set?
a) ( )
b) [ ]
c) { }
d) < >
Answer: c
Clarification: Empty set is a set which does not contain any element. It can be represented by { } or ϕ symbol.

3. Which of the following is an empty set?
a) Prime numbers up to 10
b) Even numbers up to 10
c) Prime numbers divisible by 2
d) Prime numbers divisible by 3
Answer: d
Clarification: Prime number is a number which is only divisible by 1 and itself. So, there is no prime number divisible by 3. So, it is an empty set.
Set of prime numbers up to 10 is {2,3,5,7} Set of even numbers up to 10 is {2,4,6,8} Set of prime numbers divisible by 2 is {2}.

4. Find the number of points common to parallel lines.
a) Three points
b) One point
c) Two point
d) No point
Answer: d
Clarification: Since parallel lines do not intersect anywhere they do not have any point in common. So, no point is common to parallel lines.

5. Is set {x : x is a natural number x<5 and x>7} a null set?
a) True
b) False
Answer: a
Clarification: A number less than 5 cannot be greater than 7. If we plot points less than 5 on the number line and then plot numbers greater than 7 on it we find, no point is common to both so, it is a null set.

6. Which of the following is a null set?
a) {x : x is a natural number and x2=4}
b) {x : x is a rational number and x2=2}
c) {x : x is an even prime number}
d) {x : x is name of the day of week}
Answer: b
Clarification: x2=4 => x=2,-2 where 2 is a natural number so X={2}.
X={2} for even prime number. Also we have seven days of the week so they cannot form null set.
x2=2 => x=√2, -√2 they both are irrational so, it is a null set.

7. The number of elements in a null set is ______________
a) zero
b) one
c) two
d) any
Answer: a
Clarification: Null set or empty set is a set which does not contain any element. So, the number of elements in a null set is zero.

8. Find the solution to set {x : x is a natural number and 2x+1=2}.
a) {1}
b) {2}
c) {1/2}
d) { }
Answer: d
Clarification: 2x+1=2 => 2x = 1
x=1/2 which is not a natural number. So, the solution to given set is { }.

9. A set with no elements in it is called?
a) Equivalent Set
b) Empty Set
c) Equal Set
d) Infinite Set
Answer: b
Clarification: An empty set is a set which contains 0 elements and is denoted by {}.

10. Which of the following is an empty set?
a) The set of dogs with six legs
b) The set of books in the library
c) The set of boys in a school
d) The set of a square with 4 sides
Answer: a
Clarification: Clearly there is no dog in this world with 6 legs hence this makes an empty set.

11. If B is a null set and A is some set then which one of the following is false?
a) B⊆A
b) B∪A=A
c) B∩A=A
d) B∩A=B
Answer: c
Clarification: An empty set is a subset of every set, also the union of a set with an empty set gives the set itself, Since, there are no elements in an empty set hence intersection with an empty set gives empty set.

12. The subset of a null set is the null set itself.
a) True
b) False
Answer: a
Clarification: The null set has no elements to form a subset hence its subset is a null set itself.

250+ TOP MCQs on Trigonometric Functions of Sum and Difference of Two Angles-1 & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Trigonometric Functions of Sum and Difference of Two Angles-1”.

1. cos(75°) =__________________
a) (1 – (sqrt{3}))/2(sqrt{2})
b) ((sqrt{3}) + 1)/2(sqrt{2})
c) ((sqrt{3}) – 1)/2(sqrt{2})
d) (-(sqrt{3}) – 1)/2(sqrt{2})
Answer: c
Clarification: cos(75°) = cos (45°+30°) = cos45° cos30° – sin45° sin30°
= (1/(sqrt{2}) * (sqrt{3})/2) – (1/(sqrt{2}) * 1/2) {cos(x + y)=cos x cos y – sin x sin y}
= ((sqrt{3}) – 1)/2(sqrt{2}).

2. cos(15°) =_____________
a) (1 – (sqrt{3}))/2(sqrt{2})
b) ((sqrt{3}) + 1)/2(sqrt{2})
c) ((sqrt{3}) – 1)/2(sqrt{2})
d) (-(sqrt{3}) – 1)/2(sqrt{2})
Answer: b
Clarification: cos(15°) = cos (45°-30°) = cos45° cos30° + sin45° sin30°
= (1/(sqrt{2}) * (sqrt{3})/2) + (1/(sqrt{2}) * 1/2) {cos(x – y)=cos x cos y + sin x sin y}
= ((sqrt{3}) +1)/2(sqrt{2}).

3. sin (75°) =__________________
a) (1 – (sqrt{3}))/2(sqrt{2})
b) ((sqrt{3}) + 1)/2(sqrt{2})
c) ((sqrt{3}) – 1)/2(sqrt{2})
d) (- (sqrt{3}) – 1)/2(sqrt{2})
Answer: b
Clarification: sin (75°) = sin (45°+30°) = sin45° cos30° + cos45° sin30°
= (1/(sqrt{2}) * (sqrt{3})/2) + (1/(sqrt{2}) * 1/2) {sin(x + y)=sin x cos y + cos x sin y}
= ((sqrt{3}) + 1)/2(sqrt{2}).

4. sin(15°) =_________________
a) (1 – (sqrt{3}))/2(sqrt{2})
b) ((sqrt{3}) + 1)/2(sqrt{2})
c) ((sqrt{3}) – 1)/2(sqrt{2})
d) (- (sqrt{3}) – 1)/2(sqrt{2})
Answer: c
Clarification: sin (15°) = sin (45°-30°) = sin45° cos30° – cos45° sin30°
= (1/(sqrt{2}) * (sqrt{3})/2) – (1/(sqrt{2}) * 1/2) {sin(x – y)=sin x cos y – cos x sin y}
= ((sqrt{3}) -1)/2(sqrt{2}).

5. Is cos (90° – x) = sin x.
a) True
b) False
Answer: a
Clarification: cos (90° – x) = cos 90° cos x + sin 90° sin x {cos(x – y)=cos x cos y + sin x sin y}
= 0*cos x + 1*sin x
= sin x.

6. Is sin (90°+x) = cos x.
a) True
b) False
Answer: a
Clarification: sin (90°+x) = sin 90° cos x + cos 90° sin x {sin(x + y)=sin x cos y + cos x sin y}
= 1*cos x + 0*sin x
= cos x.

7. tan(75°) =___________________
a) 2+(sqrt{3})
b) 2-(sqrt{3})
c) 1+(sqrt{3})
d) (sqrt{3})-1
Answer: a
Clarification: tan (x +y) = (tan x + tan y)/(1- tan x tan y)
tan (45°+30°) = (tan 45° + tan 30°)/(1- tan 45° tan 30°)
tan 75° = (1+ 1/(sqrt{3}))/(1-1/(sqrt{3})) = ((sqrt{3}) + 1)/ ((sqrt{3}) – 1) = 2+(sqrt{3}).

8. tan(15°) =___________________
a) 2 + (sqrt{3})
b) 2 – (sqrt{3})
c) 1 + (sqrt{3})
d) (sqrt{3}) – 1
Answer: b
Clarification: We know, tan (x -y) = (tan x – tan y)/(1+ tan x tan y)
tan (45°-30°) = (tan 45° – tan 30°)/(1+ tan 45° tan 30°)
tan 75° = (1- 1/(sqrt{3}))/ (1+ 1/(sqrt{3})) = ((sqrt{3}) – 1)/ ((sqrt{3}) + 1) = 2-(sqrt{3}).

9. cot 75° =___________________________
a) 2+(sqrt{3})
b) 2-(sqrt{3})
c) 1+(sqrt{3})
d) (sqrt{3})-1
Answer: b
Clarification: We know, cot (x +y) = (cot x cot y -1)/(cot y + cot x)
cot(45°+30°) = (cot 45° cot 30°-1)/(cot 45° + cot 30°)
cot 75° = ((sqrt{3}) – 1)/((sqrt{3}) + 1) = 2-(sqrt{3}).

10. cot 15° =______________
a) 2+(sqrt{3})
b) 2-(sqrt{3})
c) 1+(sqrt{3})
d) (sqrt{3})-1
Answer: a
Clarification: We know, cot (x – y) = (cot x cot y +1)/cot y – cot x)
cot(45°-30°) = (cot 45° cot 30°+1)/(cot 45° – cot 30°)
cot 15° = ((sqrt{3}) + 1)/((sqrt{3}) – 1) = 2+(sqrt{3}).

11. Find cos 2x if sin x=1/2.
a) 1/2
b) 1/(sqrt{2})
c) (sqrt{3})/2
d) 1
Answer: c
Clarification: We know, cos 2x = cos2x – sin2x = 1-2sin2x {cos2x = 1-sin2x}
= 1-2(1/2)2 = 1-2(1/4) = 1-1/2 = 1/2.

12. Find cos 2x if cos x = 1/(sqrt{2}).
a) 1/2
b) 0
c) (sqrt{3})/2
d) 1
Answer: b
Clarification: We know, cos 2x = cos2x – sin2x = 2cos2x – 1 {sin2x = 1-cos2x}
= 2(1/(sqrt{2}))2-1 = 2(1/2) – 1 = 1-1 = 0.

13. Find cos 2x if tan x=1/(sqrt{3}).
a) 1/2
b) 0
c) (sqrt{3})/2
d) 1
Answer: a
Clarification: We know, cos 2x = cos2x – sin2x = (cos2x – sin2x)/(cos2x+sin2x) {1 = sin2x + cos2x}
= (1-tan2x)/(1+tan2x)
= (1-(1/(sqrt{3}))2)/(1+(1/(sqrt{3}))2)
= (1-1/3)/(1+1/3) = (2/3)/(4/3) = 1/2.

250+ TOP MCQs on Permutations-2 & Answers | Class 11 Maths

Mathematics Questions and Answers for Aptitude test on “Permutations-2”.

1. Find the number of 4 letter words which can be formed from word IMAGE using permutations without repetition.
a) 20
b) 60
c) 120
d) 240
Answer: c
Clarification: We have to arrange 4 letters of the 5 letters word IMAGE without repetition. So, total permutations are nPr = 5P4 = 5! / (5-4)! = 5! = 5.4.3.2.1 = 120.

2. Find the number of 4 letter words which can be formed from word IMAGE if repetition is allowed.
a) 120
b) 125
c) 625
d) 3125
Answer: c
Clarification: We have to arrange 4 letters of the 5 letters word IMAGE without repetition. So, total permutations are nr = 54 = 625.

3. How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?
a) 504
b) 729
c) 1000
d) 720
Answer: a
Clarification: 1-9 digits which means 9 digits are possible. We have to arrange 3 digits at a time out of 9 digits without repetition. So, total permutations are nPr = 9P3 = (frac{9!}{(9-3)!} = frac{9!}{6!}) = 9.8.7 = 504.

4. How many 3-digit numbers are possible using permutations with repetition allowed if digits are 1-9?
a) 504
b) 729
c) 1000
d) 720
Answer: b
Clarification: 1-9 digits which means 9 digits are possible. We have to arrange 3 digits at a time out of 9 digits with repetition allowed. So, total permutations are nr = 93 = 729.

5. If nP3 = 4*nP2. Find n.
a) 3
b) 2
c) 6
d) 5
Answer: c
Clarification: nP3 = 4*nP2
(frac{n!}{(n-3)!} = 4 * frac{n!}{(n-2)!})
=> n-2 = 4
=> n=6.

6. 4Pr = 4*5Pr-1. Find r.
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: 4Pr = 4*5Pr-1
=> (frac{4!}{(4-r)!} = 4*frac{5!}{(5-r+1)!})
=> (frac{(6-r)!}{(4-r)!} = 4*frac{5!}{4!})
=> (6-r) (5-r) = 4*5
=> r=1.

7. Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels occur together.
a) 40320
b) 37440
c) 1440
d) 2880
Answer: d
Clarification: There are 5 vowels in word EDUCATION. 5 vowels can be arranged in 5P5 i.e. 5! ways. When all vowels are together, 5 vowels together form one letter and remaining 3 letters i.e. together 4 letters can be arranged in 4P4 i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.

8. Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels do not occur together.
a) 40320
b) 37440
c) 1440
d) 2880
Answer: b
Clarification: There are 5 vowels in word EDUCATION. 5 vowels can be arranged in 5P5 i.e. 5! ways. When all vowels are together, 5 vowels together form one letter and remaining 3 letters i.e. together 4 letters can be arranged in 4P4 i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.
So, when all vowels do not occur together, total possible arrangements = 8! – 2880 = 40320 – 2880 = 37440.

9. In how many ways 2 red pens, 3 blue pens and 4 black pens can be arranged if same color pens are indistinguishable?
a) 362880
b) 1260
c) 24
d) 105680
Answer: b
Clarification: Total number of pens are 2+3+4 = 9 out of which 2 are of 1 type, 3 are of 2nd type and 4 are of 3rd type so, total number of arrangements = (frac{9!}{2!3!4!} = frac{9*8*7*6*5}{2*6}) = 1260.

10. Find the number of words which can be made using all the letters of the word IMAGE. If these words are written as in a dictionary, what will be the rank of MAGIE?
a) 97
b) 98
c) 99
d) 100
Answer: c
Clarification: Words starting with letter A comes first in dictionary. Starting with A, number of words = 4! = 24. Starting with E, number of words = 4! = 24. Starting with I, number of words = 4! = 24. Starting with G, number of words = 4! = 24. Since our word also start with M so, we have to consider one more letter i.e. MA. Since our word also start with MA so, we have to consider one more letter i.e. MAE. Starting with MAE, number of words = 2! = 2. Since our word also start with MAG so, we have to consider one more letter i.e. MAGE. Starting with MAGE, only one letter i.e. MAGEI. After this, MAGIE comes. Total number of words before MAGIE = 24 + 24 + 24 + 24 + 2 = 98. So, rank of MAGIE is 99.

Mathematics for Aptitude test,

250+ TOP MCQs on Conic Sections – Parabola-2 & Answers | Class 11 Maths

Mathematics Quiz for Class 11 on “Conic Sections – Parabola-2”.

1. Find the equation of directrix of parabola x2=100y.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: c
Clarification: Comparing equation with x2=4ay.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is y=-a => y=-25.

2. Find the equation of directrix of parabola x2=-100y.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: d
Clarification: Comparing equation with x2=-4ay.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is y=a => y=25.

3. Find the vertex of the parabola y2=4ax.
a) (0, 4)
b) (0, 0)
c) (4, 0)
d) (0, -4)
Answer: b
Clarification: Vertex is close end point of the parabola i.e. origin (0, 0) as it satisfies the equation (y-0)2=4a(x-0) also.

4. Find the equation of axis of the parabola y2=24x.
a) x=0
b) x=6
c) y=6
d) y=0
Answer: d
Clarification: Axis is the line passing through focus which divides the parabola symmetrically into two equal halves. Equation of axis of parabola y2=24x is y=0.

5. Find the equation of axis of the parabola x2=24y.
a) x=0
b) x=6
c) y=6
d) y=0
Answer: a
Clarification: Axis is the line passing through focus which divides the parabola symmetrically into two equal halves. Equation of axis of parabola x2=24y is x=0.

6. Find the length of latus rectum of the parabola y2=40x.
a) 4 units
b) 10 units
c) 40 units
d) 80 units
Answer: c
Clarification: Comparing equation with y2=4ax.
4a=40. Length of latus rectum of parabola is 4a =40.

7. If focus of parabola is F (2, 5) and equation of directrix is x + y=2, then find the equation of parabola.
a) x2+y2+2xy-4x-16y+54=0
b) x2+y2+2xy+4x-16y+54=0
c) x2+y2+2xy+4x+16y+54=0
d) x2+y2+2xy-4x+16y+54=0
Answer: a
Clarification: We know, if P is a point on parabola, M is foot of perpendicular drawn from point P to directrix of parabola and F is focus of parabola then PF=PM
(x-2)2+(y-5)2 = (|x+y-2|/√2)2
x2+y2+2xy-4x-16y+54=0.

8. If vertex is at (1, 2) and focus (2, 0) then find the equation of the parabola.
a) y2-8x+4y+12=0
b) y2-8x-4y-12=0
c) y2-8x-4y+12=0
d) y2+8x+4y+12=0
Answer: c
Clarification: Since vertex is at (1, 2) and focus (2, 0) so parabola equation will be
(y-2)2=4*2(x-1)
y2-8x-4y+12=0.

9. Equation of parabola which is symmetric about x-axis with vertex (0, 0) and pass through (3, 6).
a) y2=6x
b) x2=12y
c) y2=12x
d) x2=6y
Answer: c
Clarification: Equation of parabola which is symmetric about x-axis with vertex (0, 0) is y2=4ax
Since parabola pass through (3, 6) then 62=4a*3 => a=3.
So, equation is y2=12x.

10. Equation of parabola which is symmetric about y-axis with vertex (0, 0) and pass through (6, 3).
a) y2=6x
b) x2=12y
c) y2=12x
d) x2=6y
Answer: b
Clarification: Equation of parabola which is symmetric about y-axis with vertex (0, 0) is x2=4ay
Since parabola pass through (6, 3) then 62=4a*3 => a=3.
So, equation is x2=12y.