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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 1-30 of 356 questions
1 Easy
1 of 356
Application +4 | -1
The equation of a common tangent to the parabolas y=x2y=x^2 and y=−(x−2)2y=-(x-2)^2 is
View answer and explanation
Correct option: B
y=4(x−1)y=4(x-1)
Option B is correct because the common tangent is found by equating the tangent equations of both parabolas and solving for the slope.
2 Easy
2 of 356
Application +4 | -1
A point PP moves such that slope of one tangent drawn from it to parabola y2=16xy^2=16 x is four times the slope of other tangent, then the length of latus rectum of the conic represented by locus of PP, is
View answer and explanation
Correct option: B
25
Option B (25) is correct because the locus of point P is a parabola, and the length of its latus rectum is calculated to be 25.
3 Easy
3 of 356
Analyzing +4 | -1
The tangents at the end points of any chord through (1,0)(1,0) to the parabola y2+4x=8y^2+4 x=8 intersect
View answer and explanation
Correct option: D
at 90∘90^{\circ} on x−3=0x-3=0
Option D (at 90∘90^{\circ} on x−3=0x-3=0) is correct because the tangents at the endpoints of a focal chord intersect at right angles on the directrix.
4 Easy
4 of 356
Application +4 | -1
The focus of the parabola x2=−16yx^2=-16 y is
View answer and explanation
Correct option: D
(0,−4)(0,-4)
Option D ((0,−4)(0,-4)) is correct because for the parabola x2=−16yx^2 = -16y, the focus is at (0, -a) where a = 4.
5 Easy
5 of 356
Application +4 | -1
The equation of the parabola with its vertex at the origin, axis on the y−axisy - axis and passing through the point (6,−3)\left( {6, - 3} \right) is
View answer and explanation
Correct option: C
x2=−12y{x^2} = - 12y
Option C (x2=−12y{x^2} = - 12y) is correct because it is the equation of a parabola with the given vertex, axis, and passing point.
6 Easy
6 of 356
Application +4 | -1
Let L1L_1 be a tangent to the parabola y2=4(x+1)y^2=4(x+1) and L2L_2 be a tangent to the parabola y2=8(x+2)y^2=8(x+2) such that L1L_1 and L2L_2 intersect at right angles. Then L1L_1 and L2L_2 meet on the straight line
View answer and explanation
Correct option: A
x+3=0x+3=0
Option A (x+3=0x+3=0) is correct because the intersection point of the perpendicular tangents lies on the directrix of the equivalent parabola.
7 Easy
7 of 356
Application +4 | -1
The conditions that ax+by+c=0ax + by + c = 0 is tangent to the parabola y2=4ax{y^2} = 4ax, is
View answer and explanation
Correct option: C
b2=c{b^2} = c
Option C (b2=c{b^2} = c) is correct because it represents the condition for the line ax+by+c=0ax + by + c = 0 to be tangent to the parabola y2=4ax{y^2} = 4ax.
8 Easy
8 of 356
Application +4 | -1
The point P(9/2,6)P\left( {9/2,6} \right) lies on the parabola y2=4ax{y^2} = 4ax, then parameter of the point PP is
View answer and explanation
Correct option: D
32\frac{3}{2}
Option D (32\frac{3}{2}) is correct because it is the parameter of the point P on the given parabola.
9 Easy
9 of 356
Application +4 | -1
The equation of the tangent to the parabola y2=8x{y^2} = 8x which is perpendicular to the line x−3y+8=0x - 3y + 8 = 0 is
View answer and explanation
Correct option: A
9x+3y+2=09x + 3y + 2 = 0
Option A (9x+3y+2=09x + 3y + 2 = 0) is correct because it is found by using the formula for the tangent to a parabola when the slope is known.
10 Easy
10 of 356
Application +4 | -1
The focus of the parabola y2−4y−8x+4=0y^2-4 y-8 x+4=0, is
View answer and explanation
Correct option: D
(2,2)(2,2)
Option D ((2,2)) is correct because completing the square and rewriting the equation reveals the vertex form, allowing us to determine the focus.
11 Easy
11 of 356
Application +4 | -1
The equation of the directrix of the parabola x2+8y−2x=7{x^2} + 8y - 2x = 7 is
View answer and explanation
Correct option: A
y=3y = 3
Option A (y=3y = 3) is correct because after rewriting the parabola equation in standard form, the directrix can be found using the vertex and the focal length.
12 Easy
12 of 356
Application +4 | -1
The conic represented by the equation x2+y2−2xy+20x+10=0x^2+y^2-2 x y+20 x+10=0, is
View answer and explanation
Correct option: C
Parabola
Option C (Parabola) is correct because the discriminant Δ≠0\Delta \neq 0 and h2=abh^2 = ab, which are the conditions for a parabola.
13 Easy
13 of 356
Application +4 | -1
The point of intersection of the parabola at the points t1t_1 and t2t_2 is
View answer and explanation
Correct option: A
[at1t2,a(t1+t2)]\left[a t_1 t_2, a\left(t_1+t_2\right)\right]
Option A ([at_1t_2, a(t_1+t_2)]) is correct because it is the standard formula for the intersection point of tangents to a parabola at points t1 and t2.
14 Easy
14 of 356
Application +4 | -1
The locus of the poles of the focal chords of a parabola is …. of the parabola
View answer and explanation
Correct option: C
The directrix
Option C (The directrix) is correct because the locus of the poles of focal chords of a parabola is its directrix.
15 Easy
15 of 356
Analyzing +4 | -1
The equation of a tangent to the parabola y2=8xy^2=8 x is y=x+2y=x+2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent, is
View answer and explanation
Correct option: B
(−2,0)(-2,0)
Option B ((-2,0)) is correct because the locus of the intersection of perpendicular tangents is the directrix, and the point lies on both the tangent and the directrix.

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16 Easy
16 of 356
Application +4 | -1
The equation of latusrectum of a parabola is x+y=8x + y = 8 and the equation of the tangent at the vertex is x+y=12,x + y = 12, then length of the latusrectum is
View answer and explanation
Correct option: D
828\sqrt 2
Option D (828\sqrt 2) is correct because it is calculated from the distance between the latus rectum and the tangent at the vertex of the parabola.
17 Easy
17 of 356
Application +4 | -1
The equation of the parabola with its vertex at (1,1)(1,1) and focus at (3,1)(3,1), is
View answer and explanation
Correct option: B
(y−1)2=8(x−1)(y-1)^2=8(x-1)
Option B ((y−1)2=8(x−1)(y-1)^2=8(x-1)) is correct because it represents a parabola with the given vertex and focus, opening along the positive x-axis.
18 Easy
18 of 356
Application +4 | -1
If the line y=2x+ky=2 x+k is a tangent to the curve x2=4yx^2=4 y, then kk is equal to
View answer and explanation
Correct option: C
-4
Option C (-4) is correct because the condition for tangency is c=−am2c = -a m^2.
19 Easy
19 of 356
Analyzing +4 | -1
If a chord which is normal to the parabola y2=4ax{y^2} = 4ax at one end subtends a right angle at the vertex, then its slope is
View answer and explanation
Correct option: C
2\sqrt 2
Option C (2\sqrt 2) is correct because it is the slope of the normal that subtends a right angle at the vertex of the parabola, derived by setting the sum of coefficients of x² and y² to zero.
20 Easy
20 of 356
Application +4 | -1
Let one end of a focal chord of parabola y2=8xy^2=8 x be at (1/2,−2)(1 / 2,-2). Then the equation of the tangent at the other end of this focal chord is
View answer and explanation
Correct option: D
x−2y+8=0x-2 y+8=0
Option D (x−2y+8=0x-2 y+8=0) is correct because it represents the equation of the tangent at the other end of the focal chord, calculated using the property t1t2=−1t_1 t_2 = -1.
21 Easy
21 of 356
Application +4 | -1
The locus of mid-point of the line segment joining the focus to a moving point on the parabola y2=4axy^2=4 a x is another parabola with directrix
View answer and explanation
Correct option: C
x=0x=0
Option C (x=0) is correct because the locus of the midpoint is another parabola, and its directrix is found to be x=0.
22 Easy
22 of 356
Application +4 | -1
If x2+6x+20y−51=0{x^2} + 6x + 20y - 51 = 0 , then axis of parabola is
View answer and explanation
Correct option: A
x+3=0x + 3 = 0
Option A (x+3=0x + 3 = 0) is correct because it represents the axis of the given parabola after completing the square.
23 Easy
23 of 356
Application +4 | -1
The two ends of latusrectum of a parabola are the points (3,6)\left( {3,6} \right) and (−5,6).\left( { - 5,6} \right). The focus is
View answer and explanation
Correct option: B
(−1,6)\left( { - 1,6} \right)
Option B ((-1,6)) is correct because the focus is the midpoint of the latus rectum.
24 Easy
24 of 356
Analyzing +4 | -1
If the line y=4+kx,    k>0,y = 4 + kx,\,\,\,\,k > 0, is the tangent to the parabola y=x−x2y = x - {x^2} at the point P and V is the vertex of the parabola, then the slope of the line through P and V is:
View answer and explanation
Correct option: C
05-Feb
The correct answer is 05-Feb because the slope of the line through P and V is calculated by finding the point of tangency P and the vertex V of the parabola and then calculating the slope between these two points, resulting in 5/2.
25 Easy
25 of 356
Application +4 | -1
The locus of the point of intersection of the perpendicular tangents to the parabola x2=4ayx^2=4 a y is
View answer and explanation
Correct option: B
Directrix of the parabola
Option B (Directrix of the parabola) is correct because the locus of the point of intersection of perpendicular tangents to a parabola is its directrix.
26 Easy
26 of 356
Application +4 | -1
If (0,4)(0,4) and (0,2)(0,2) are respectively the vertex and focus of a parabola, then its equation, is
View answer and explanation
Correct option: A
x2+8y=32x^2+8 y=32
Option A (x2+8y=32x^2+8y=32) is correct because it represents a parabola with the given vertex and focus, opening downwards.
27 Easy
27 of 356
Application +4 | -1
The equation of the parabola whose vertex and focus lie on the axis of xx at distances aa and a1a_1 from the origin respectively, is
View answer and explanation
Correct option: B
y2=4(a1−a)(x−a)y^2=4\left(a_1-a\right)(x-a)
Option B (y2=4(a1−a)(x−a)y^2=4(a_1-a)(x-a)) is correct because it represents a parabola with vertex at (a,0) and focus at (a1,0).
28 Easy
28 of 356
Application +4 | -1
In the parabola y2=6xy^2=6 x, the equation of the chord through the vertex and the negative end of the latus rectum, is
View answer and explanation
Correct option: B
y+2x=0y+2 x=0
Option B (y+2x=0y+2 x=0) is correct because it represents the equation of the line passing through the vertex (0,0) and the negative end of the latus rectum.
29 Easy
29 of 356
Application +4 | -1
If P(at2,  2at)P\left( {a{t^2},\;2at} \right) be one end of a focal chord of the parabola y2=4ax{y^2} = 4ax, then the length of the chord is
View answer and explanation
Correct option: D
a(t+1t)2a{\left( {t + \frac{1}{t}} \right)^2}
Option D (a(t+1t)2a{\left( {t + \frac{1}{t}} \right)^2}) is correct because it represents the length of the focal chord of the parabola.
30 Easy
30 of 356
Application +4 | -1
If the vertex of a parabola be at origin and directrix be x+5=0x+5=0, then its latus rectum is
View answer and explanation
Correct option: C
20
Option C (20) is correct because the latus rectum of a parabola is 4a, where 'a' is the distance from the vertex to the directrix.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 356 single-correct-answer MCQs on PARABOLA.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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