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JEE • MATHEMATICS • PARABOLA

Parabola practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-90 of 356 questions

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61 Easy
61 of 356
Application +4 | -1
The equation of the directrix of the parabola x2−4x−3y+10=0,{x^2} - 4x - 3y + 10 = 0, is
View answer and explanation
Correct option: B
y=54y = \frac{5}{4}
Option B (y=54y = \frac{5}{4}) is correct because it represents the directrix of the given parabola after rewriting it in standard form.
62 Easy
62 of 356
Analyzing +4 | -1
Let A,B,CA, B, C be three points on the parabola y2=4axy^2=4 a x such that tangents at these points taken in pairs form a triangle PQRP Q R. Then, area (△ABC)(\triangle A B C) : area (△PQR)=(\triangle P Q R)=
View answer and explanation
Correct option: B
2:12: 1
Option B (2:1) is correct. The ratio of the area of the triangle formed by three points on a parabola to the area of the triangle formed by the tangents at those points is always 2:1.
63 Easy
63 of 356
Application +4 | -1
The point of intersection of the normals to the parabola y2=4xy^{2}=4 x at the ends of its latus rectum is :
View answer and explanation
Correct option: B
(3,0)(3,0)
Option B ((3,0)(3,0)) is correct because the point of intersection of the normals to the parabola y2=4axy^2 = 4ax at the ends of its latus rectum is (3a, 0).
64 Easy
64 of 356
Application +4 | -1
Vertex of the parabola is (2,−1)(2,-1) and equation of the directrix is 4x−3y=21.4 x-3 y=21. Then length of the latusrectum is
View answer and explanation
Correct option: B
8
Option B (8) is correct because the length of the latus rectum of a parabola is 4a, where a is the distance from the vertex to the directrix, and in this case, 4a = 8.
65 Easy
65 of 356
Application +4 | -1
If a chord which is not a tangent, of the parabola y2=16xy^2=16 x has the equation 2x+y=p2 x+y=p, and mid-point (h,k)(h, k), then which of the following is (are) possible value (s) of p,hp, h and kk ?
View answer and explanation
Correct option: D
p=2,h=3,k=−4p=2, h=3, k=-4
Option D (p=2,h=3,k=−4p=2, h=3, k=-4) is correct because these values satisfy the equation derived from the chord's midpoint and the parabola's equation.
66 Easy
66 of 356
Application +4 | -1
If the vertex of the parabola y=x2−16x+ky = {x^2} - 16x + k lies on xx-axis, then the value of kk is
View answer and explanation
Correct option: C
64
Option C (64) is correct because when k=64, the equation becomes a perfect square, indicating the vertex lies on the x-axis.
67 Easy
67 of 356
Analyzing +4 | -1
Consider the two curves
C1:y2=4x,C2:x2+y2−6x+1=0{C_1}:{y^2} = 4x,{C_2}:{x^2} + {y^2} - 6x + 1 = 0
Then
View answer and explanation
Correct option: B
C1C_1 and C2C_2 touch each other exactly at two point
Option B is correct. The curves C1C_1 and C2C_2 touch each other exactly at two points, which are (1, 2) and (1, -2), as they have the same tangent at these points.
68 Easy
68 of 356
Analyzing +4 | -1
The slope of a chord of the parabola y2=4axy^2=4 a x which is normal at one end and which subtends a right angle at the origin, is
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because the slope of the normal chord is ±2\pm \sqrt{2} when it subtends a right angle at the origin.
69 Easy
69 of 356
Application +4 | -1
If tangents at extremities of a focal chord ABAB of the parabola y2=4ax{y^2} = 4ax intersect at a point CC, then ∠ACB\angle ACB is equal to
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi }{2}) is correct because the tangents at the extremities of a focal chord of a parabola intersect at right angles.
70 Easy
70 of 356
Analyzing +4 | -1
Let x=2t,  y=t23x = 2t,\,\,y = \frac{{{t^2}}}{3} be a conic. Let SS be the focus and BB be the point on the axis of the conic such that SA⊥BASA \bot BA where AA is any point on the conic. If kk is the ordinate of the centroid of the ΔSAB,\Delta SAB, then lim⁡t→1k \lim\limits_{t \to 1} k\, is equal to
View answer and explanation
Correct option: D
1318\frac{{13}}{{18}}
The correct answer is D (1318\frac{{13}}{{18}}) because it is the limit of the ordinate of the centroid of the triangle SAB as t approaches 1, calculated using the properties of the conic and the given geometric conditions.
71 Easy
71 of 356
Analyzing +4 | -1
The number of distinct normals that can be drawn from (−2,1)(-2,1) to the parabola y2−4x−2y−3=0y^2-4 x-2 y-3=0, is
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one normal can be drawn from the point (-2, 1) to the given parabola.
72 Easy
72 of 356
Application +4 | -1
The centre of the conic 14x2−4xy+11y2−44x−58y+71=0,14 x^2-4 x y+11 y^2-44 x-58 y+71=0, is
View answer and explanation
Correct option: A
(2,3)(2,3)
Option A (2,3) is correct because it is found by solving the partial derivatives of the conic equation with respect to x and y.
73 Easy
73 of 356
Application +4 | -1
The coordinates of the extremities of the latus rectum of the parabola 5y2=4x5 y^2=4 x are
View answer and explanation
Correct option: B
(1/5,2/5),(1/5,−2/5)(1 / 5,2 / 5),(1 / 5,-2 / 5)
Option B ((1/5,2/5),(1/5,−2/5)(1 / 5,2 / 5),(1 / 5,-2 / 5)) is correct because these are the coordinates of the endpoints of the latus rectum for the given parabola.
74 Easy
74 of 356
Application +4 | -1
If the common tangent to the parabolas y2=4xy^2=4 x and x2=4yx^2=4 y also touches the circle x2+y2=c2x^2+y^2=c^2, then cc is equal to
View answer and explanation
Correct option: A
12\frac{1}{\sqrt{2}}
Option A (12\frac{1}{\sqrt{2}}) is correct because the common tangent's distance from the origin equals the radius of the circle, leading to c = 12\frac{1}{\sqrt{2}}.
75 Easy
75 of 356
Application +4 | -1
The area of the triangle inscribed in the parabola y2=4x{y^2} = 4x the ordinates of whose vertices are 1,2 and 4 is
View answer and explanation
Correct option: D
34\frac{3}{4} sq. units
Option D (34\frac{3}{4} sq. units) is correct because it is the area of the triangle calculated using the formula for the area of a triangle inscribed in a parabola.
76 Easy
76 of 356
Application +4 | -1
The equation of the parabola whose vertex and focus are on the positive side of the xx-axis at distances aa and bb respectively from the origin, is
View answer and explanation
Correct option: A
y2=4(b−a)(x−a)y^2=4(b-a)(x-a)
Option A (y2=4(b−a)(x−a)y^2=4(b-a)(x-a)) is correct because it represents a parabola with the given vertex and focus, with the correct latus rectum length.
77 Easy
77 of 356
Application +4 | -1
If (2,0)(2,0) is the vertex and yy-axis the directrix of a parabola, then its focus is
View answer and explanation
Correct option: C
(4,0)(4,0)
Option C ((4,0)(4,0)) is correct because the focus is equidistant from the vertex as the directrix, but on the opposite side.
78 Easy
78 of 356
Application +4 | -1
PQPQ is any focal chord of the parabola y2=32xy^2=32 x. The length of PQP Q can never be less than
View answer and explanation
Correct option: C
32 units
Option C (32 units) is correct because the shortest focal chord is the latus rectum, and for the parabola y2=4axy^2 = 4ax, the length of the latus rectum is 4a4a.
79 Easy
79 of 356
Application +4 | -1
The equation of the parabola whose vertex is (−1,−2)(-1,-2), axis is vertical, and which passes through the point (3,6)(3,6), is
View answer and explanation
Correct option: A
x2+2x−2y−3=0x^2+2 x-2 y-3=0
Option A (x2+2x−2y−3=0x^2+2 x-2 y-3=0) is correct because substituting the vertex and the given point into the parabola equation yields this result.
80 Easy
80 of 356
Application +4 | -1
If the line x+y−1=0x + y - 1 = 0 is a tangent to the parabola y2−y+x=0{y^2} - y + x = 0, then the point of contact is
View answer and explanation
Correct option: A
(0,1)\left( {0,1} \right)
Option A ((0,1)) is correct because it's the only point that satisfies both the line and parabola equations, indicating it's the point of contact.
81 Easy
81 of 356
Application +4 | -1
The equation of a common tangent to y2=4xy^2=4 x and the curve x2+4y2=8x^2+4 y^2=8 can be
View answer and explanation
Correct option: B
x+2y+4=0x+2 y+4=0
Option B (x+2y+4=0x+2 y+4=0) is correct because it is a common tangent to both the parabola and the ellipse, found by comparing the tangent equations of both curves.
82 Easy
82 of 356
Application +4 | -1
Two perpendicular tangents PA{PA} and PB{PB} are drawn to y2=4ax{y}^2=4 {ax}, minimum length of AB{AB} is equal to
View answer and explanation
Correct option: B
4a4 a
Option B (4 a) is correct because the chord of contact of mutually perpendicular tangents to a parabola is a focal chord, and the minimum length of a focal chord is the latus rectum, which is 4a.
83 Easy
83 of 356
Application +4 | -1
The ends of a line segment are P(1,  3)P\left( {1,\;3} \right) and Q(1,  1).Q\left( {1,\;1} \right). RR is a point on the line segment PQPQ such that PR:QR=1:λPR:QR = 1:\lambda. If RR is an interior point of the parabola y2=4x{y^2} = 4x, then
View answer and explanation
Correct option: A
λ∈(0,  1)\lambda \in \left( {0,\;1} \right)
Option A (λ∈(0,  1)\lambda \in \left( {0,\;1} \right)) is correct because the point R lies inside the parabola only when the ratio λ\lambda is between 0 and 1.
84 Easy
84 of 356
Application +4 | -1
The axis of the parabola 9y2−16x−12y−57=09 y^2-16 x-12 y-57=0, is
View answer and explanation
Correct option: A
3y=23 y=2
Option A (3y=23y=2) is correct because it represents the axis of symmetry for the given parabola after completing the square.
85 Easy
85 of 356
Application +4 | -1
If x=my+cx = my + c is a normal to the parabola x2=4  ay,{x^2} = 4\;ay, then the value of cc is
View answer and explanation
Correct option: A
−2  am−am3- 2\;am - a{m^3}
Option A (−2  am−am3- 2\;am - a{m^3}) is correct because this is the standard form for the constant term in the normal equation to the given parabola.
86 Easy
86 of 356
Application +4 | -1
The equation of the common tangent to the parabolas y2=4axy^2=4ax and x2=4byx^2=4 b y, is
View answer and explanation
Correct option: B
a1/3x+b1/3y+(ab)2/3=0a^{1 / 3} x+b^{1 / 3} y+(a b)^{2 / 3}=0
Option A (a1/3x+b1/3y+(ab)2/3=0a^{1 / 3} x+b^{1 / 3} y+(a b)^{2 / 3}=0) is correct because it is derived by equating the tangent equations of the two parabolas and solving for the common tangent.
87 Easy
87 of 356
Application +4 | -1
The line x+y=6x+y=6 is a normal to the parabola y2=8xy^2=8 x at the point.
View answer and explanation
Correct option: C
(2,4)(2,4)
Option C ((2,4)) is correct because the slope of the normal is -1, and substituting this into the parabola's normal equation gives the point (2,4).
88 Easy
88 of 356
Analyzing +4 | -1
Consider the parabola y2=8xy^2=8 x. Let Δ1\Delta_1 be the area of the triangle formed by the end points of its latus rectum and the point P(1/2,2)P(1 / 2,2) on the parabola, and Δ2\Delta_2 be the area of the triangle formed by drawing tangents at PP and at the end-points of the latus rectum. Then, Δ1Δ2\frac{\Delta_1}{\Delta_2}, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the area of a triangle formed by three points on a parabola is twice the area of the triangle formed by tangents at those points.
89 Easy
89 of 356
Application +4 | -1
The focus of the parabola y=2x2+xy = 2{x^2} + x is
View answer and explanation
Correct option: C
(−14,0)\left( { - \frac{1}{4},0} \right)
Option D ((-14\frac{1}{4},0)) is correct because it represents the focus of the given parabola after completing the square and transforming the equation.
90 Easy
90 of 356
Analyzing +4 | -1
The normals at the ends of the latusrectum of the parabola y2=4axy^2=4 a x intersect at angle of
View answer and explanation
Correct option: D
π2\frac{\pi}{2}
Option D (π2\frac{{\pi}}{{2}}) is correct because the normals at the ends of the latus rectum are perpendicular to each other.
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