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JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

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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 586 questions
1 Easy
1 of 586
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 586
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 Medium
3 of 586
Analyzing +4 | -1
Let P,Q,RP,Q,R be three points on parabola y2=4x{y^2} = 4x and normal at PP and RR meet at QQ, then the locus of the mid-point of the chord PRPR is a parabola whose vertex is at
View answer and explanation
Correct option: C
(−2,0)\left( { - 2,0} \right)
Option C ((−2,0)\left( { - 2,0} \right)) is correct because it is the vertex of the parabola that represents the locus of the midpoint of the chord PR.
4 Easy
4 of 586
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.
5 Easy
5 of 586
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.
6 Hard
6 of 586
Application +4 | -1
If II denotes the semi-latusrectum of the parabola y2=4ax{y^2} = 4ax and SPSP and SQSQ denote the segments of any focal chord PQ,SPQ,S being the focus, then SP,lSP,l and SQSQ are in the relation
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the semi-latus rectum ll, and the segments of the focal chord SPSP and SQSQ satisfy the harmonic progression relation, i.e., 2/l=1/SP+1/SQ2/l = 1/SP + 1/SQ.
7 Easy
7 of 586
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.
8 Medium
8 of 586
Application +4 | -1
The ordinates of the triangle inscribed in parabola y2=4axy^2=4 a x are y1,y2y_1, y_2 and y3y_3. Then the area of the triangle is
View answer and explanation
Correct option: C
18a(y1−y2)(y2−y3)(y3−y1)\frac{1}{8 a}\left(y_1-y_2\right)\left(y_2-y_3\right)\left(y_3-y_1\right)
Option C (18a(y1−y2)(y2−y3)(y3−y1)\frac{1}{8 a}(y_1-y_2)(y_2-y_3)(y_3-y_1)) is correct because the area of the triangle inscribed in the parabola is calculated using this formula.
9 Medium
9 of 586
Application +4 | -1
Let PQ be a focal chord of the parabola y2=4axy^2=4 a x If the tangents to the parabola at PP and QQ meet at a point lying on the line y=2x+a,a>0y=2 x+a, a>0, then the length of chord PQP Q, is
View answer and explanation
Correct option: B
5a5 a
Option B (5a5a) is correct because, with t1t2=−1t_1t_2 = -1 and the intersection point on y=2x+ay = 2x + a, we find t1+t2=−1t_1 + t_2 = -1, leading to PQ=5aPQ = 5a.
10 Medium
10 of 586
Application +4 | -1
The common tangent of the parabolas y2=4xy^2=4 x and x2=−8yx^2=-8 y, is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the calculated common tangent equation doesn't match any of the given options.
11 Easy
11 of 586
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.
12 Medium
12 of 586
Analyzing +4 | -1
If the chord of contact of tangents from a point PP to the parabola y2=4ax{y^2} = 4ax touches the parabola x2=4by,{x^2} = 4by, then the locus of PP is a/an
View answer and explanation
Correct option: D
Hyperbola
Option D (Hyperbola) is correct because the locus of P is a hyperbola, derived from the condition that the chord of contact touches another parabola.
13 Medium
13 of 586
Application +4 | -1
Let PQP Q be a focal chord of the parabola y2=4xy^2=4 x. If the centre of a circle having PQP Q as its diameter lies on the line 5y+4=0\sqrt{5} y+4=0, then length of the chord PQP Q, is
View answer and explanation
Correct option: A
365\frac{36}{5}
Option A (36/5) is correct because the length of the focal chord can be determined using the given condition on the center of the circle.
14 Easy
14 of 586
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.
15 Hard
15 of 586
Application +4 | -1
A common tangent is drawn to the circle x2+y2=c2x^2+y^2=c^2 and the parabola y2=4axy^2=4 a x. If the angle which this tangent makes with the axis of xx is π/4\pi / 4 then the relationship between a and c(a,c>0)c(a, c>0) is-
View answer and explanation
Correct option: A
a=2c\mathrm{a}=\sqrt{2} \mathrm{c}
Option D (2x+y−33=02 x+y-3 \sqrt{{3}}=0) is correct because it is the equation of the tangent calculated at the point where the ordinate and abscissa are equal on the curve.

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16 Medium
16 of 586
Application +4 | -1
If a point PP moves such that its distances from the point A(1,1)A\left( {1,1} \right) and the line x+y+2=0x + y + 2 = 0 are equal, then the locus of PP is
View answer and explanation
Correct option: C
A parabola
Option C (A parabola) is correct because the given condition leads to an equation that represents a parabola.
17 Easy
17 of 586
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.
18 Easy
18 of 586
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.
19 Medium
19 of 586
Application +4 | -1
Tangents at P(t1)P\left( {{t_1}} \right) and Q(t2)Q\left( {{t_2}} \right) on the curve y2=4ax{y^2} = 4ax are meeting at a point RR on the axis of the parabola. the area of Δ PQR\Delta \,PQR is
View answer and explanation
Correct option: C
4a2t1t224{a^2}{t_1}t_2^2
Option D (None of these) is incorrect. The correct answer is 4a2t12t24{a^2}{t_1}^2{t_2}
20 Easy
20 of 586
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.
21 Medium
21 of 586
Analyzing +4 | -1
The circle x2+y2−2x−6y+2=0x^2+y^2-2 x-6 y+2=0 intersects the parabola y2=8xy^2=8 x orthogonally at the point PP. The equation of the tangent to the parabola at PP can be
View answer and explanation
Correct option: D
2x−y+1=02 x-y+1=0
Option D (2x−y+1=02x - y + 1 = 0) is correct because it represents a tangent to the parabola that passes through the center of the circle, ensuring orthogonality.
22 Medium
22 of 586
Analyzing +4 | -1
If P(−3,2)P\left( { - 3,2} \right) is one end of the focal chord PQPQ of the parabola y2+4x+4y=0,{y^2} + 4x + 4y = 0, then the slope of the normal at QQ is
View answer and explanation
Correct option: A
−1/2- 1/2
Option A (-1/2) is correct because the tangent at one end of a focal chord is parallel to the normal at the other end.
23 Easy
23 of 586
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.
24 Medium
24 of 586
Application +4 | -1
The condition for which the straight line y=mx+cy=m x+c touches the parabola y2=4axy^2=4 a x is
View answer and explanation
Correct option: B
ac=m\frac{a}{c}=m
Option B (ac=m\frac{{a}}{{c}}=m) is correct because it represents the condition for the line y=mx+cy=mx+c to be tangent to the parabola y2=4axy^2=4ax.
25 Medium
25 of 586
Analyzing +4 | -1
The set of points on the axis of the parabola y2−2y−4x+5=0{y^2} - 2y - 4x + 5 = 0 from which all the three normals to the parabola are real, is
View answer and explanation
Correct option: A
{(x,1):x≥3}\left\{ {\left( {x,1} \right):x \ge 3} \right\}
Option A ({(x,1):x≥3}\left\{ {\left( {x,1} \right):x \ge 3} \right\}) is correct because it defines the set of points on the axis of the parabola from which all three normals to the parabola are real.
26 Medium
26 of 586
Application +4 | -1
Equation of common tangents of the parabola y2=4xy^2=4 x and x2=4yx^2=4 y.
View answer and explanation
Correct option: A
y=−x−1y=-x-1
Option A (y=-x-1) is correct because it is the common tangent to both parabolas, found by equating the tangent equations and solving for the slope.
27 Easy
27 of 586
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.
28 Easy
28 of 586
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.
29 Medium
29 of 586
Application +4 | -1
The line y=mx+1y = mx + 1 is a tangent to the parabola y2=4x{y^2} = 4x, if
View answer and explanation
Correct option: A
m=1m = 1
Option A (m = 1) is correct because substituting the tangent equation into the parabola equation and solving for tangency condition yields m = 1.
30 Easy
30 of 586
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.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 586 single-correct-answer MCQs on PARABOLA.
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