Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Parabola question

2019 · 9 Jan · Shift 1 · Q33
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Parabola
  5. /2019 · 9 Jan · Shift 1 · Q33

Parabola question

2019 · 9 Jan · Shift 1 · Q33

JEE MainMathematicsParabolaMCQ+4 / −1
Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin, on the positive x-axis then which of the following points does not lie on it?
  1. A
    (5, 2 6\sqrt 66​)
  2. B
    (6, 4 2\sqrt 22​)
  3. C
    (8, 6)
  4. D
    (4, -4)
View written solutionFree

Correct answer: C

  1. Identify the parabola

Since the axis lies along the positive xxx-axis, and the vertex and focus are on the positive xxx-axis at distances 222 and 444 from the origin respectively, we have:

  • Vertex V=(2,0)V = (2,0)V=(2,0)
  • Focus F=(4,0)F = (4,0)F=(4,0)

So the parabola opens to the right.

The standard form of a parabola opening right with vertex (h,k)(h,k)(h,k) is:

(y−k)2=4a(x−h)(y-k)^2 = 4a(x-h)(y−k)2=4a(x−h)

where focus is (h+a,k)(h+a,k)(h+a,k).

Here,

a=4−2=2a = 4-2 = 2a=4−2=2

Hence the equation is:

y2=4(2)(x−2)=8(x−2)y^2 = 4(2)(x-2) = 8(x-2)y2=4(2)(x−2)=8(x−2)

So the parabola is

y2=8(x−2)y^2 = 8(x-2)y2=8(x−2)
  1. Check each option

We test whether each point satisfies y2=8(x−2)y^2 = 8(x-2)y2=8(x−2).

Option A: (5,26)(5, 2\sqrt 6)(5,26​)

LHS:

y2=(26)2=24y^2 = (2\sqrt 6)^2 = 24y2=(26​)2=24

RHS:

8(x−2)=8(5−2)=248(x-2) = 8(5-2) = 248(x−2)=8(5−2)=24

So option A lies on the parabola.


Option B: (6,42)(6, 4\sqrt 2)(6,42​)

LHS:

y2=(42)2=32y^2 = (4\sqrt 2)^2 = 32y2=(42​)2=32

RHS:

8(x−2)=8(6−2)=328(x-2) = 8(6-2) = 328(x−2)=8(6−2)=32

So option B lies on the parabola.


Option C: (8,6)(8, 6)(8,6)

LHS:

y2=62=36y^2 = 6^2 = 36y2=62=36

RHS:

8(x−2)=8(8−2)=488(x-2) = 8(8-2) = 488(x−2)=8(8−2)=48

Since 36≠4836 \ne 4836=48, option C does not lie on the parabola.


Option D: (4,−4)(4,-4)(4,−4)

LHS:

y2=(−4)2=16y^2 = (-4)^2 = 16y2=(−4)2=16

RHS:

8(x−2)=8(4−2)=168(x-2) = 8(4-2) = 168(x−2)=8(4−2)=16

So option D lies on the parabola.


  1. Conclusion

The point that does not lie on the parabola is:

C (8,6)\boxed{\text{C } (8,6)}C (8,6)​
  1. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They match.

PreviousNext

More from Parabola

  • If θ denotes the acute angle between the curves, y = 10 – x2 and y = 2 + x2 at a point of their intersection, the |tan θ| is equal to :2019 · MCQ
  • Let A(4, − 4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of Δ ACB is maximum. Then, the area (in sq. units) of Δ ACB, is :2019 · MCQ
  • The length of the chord of the parabola x2 = 4y having equation x –2​y+42​=0 is -2019 · MCQ
  • If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :2019 · MCQ
  • The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is :2019 · MCQ
  • Let P(4, –4) and Q(9, 6) be two points on the parabola, y2 = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of Δ PXQ is maximum. Then this maximum area (in sq.…2019 · MCQ
  • Tangents drawn from the point (− 8, 0) to the parabola y2 = 8x touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :2018 · MCQ
  • Let O be the vertex and Q be any point on the parabola, x2=8y. If the point P divides the line segment OQ internally in the ratio 1:3, then locus of P is :2015 · MCQ