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JEE • MATHEMATICS • PARABOLA

Parabola practice · Medium

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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 211 questions
1 Medium
1 of 211
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.
2 Medium
2 of 211
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.
3 Medium
3 of 211
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.
4 Medium
4 of 211
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.
5 Medium
5 of 211
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.
6 Medium
6 of 211
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.
7 Medium
7 of 211
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.
8 Medium
8 of 211
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}
9 Medium
9 of 211
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.
10 Medium
10 of 211
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.
11 Medium
11 of 211
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.
12 Medium
12 of 211
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.
13 Medium
13 of 211
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.
14 Medium
14 of 211
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.
15 Medium
15 of 211
Application +4 | -1
The equation to a parabola which passes through the intersection of a straight line x+y=0x+y=0 and the circle x2+y2+4y=0x^2+y^2+4 y=0 is
View answer and explanation
Correct option: C
y2=2xy^2=2 x
Option C (y2=2xy^2=2x) is correct because the parabola passes through the intersection points of the line and the circle.

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16 Medium
16 of 211
Application +4 | -1
If tt is the parameter for one end of a focal chord of the parabola y2=4ax,{y^2} = 4ax, then its length is
View answer and explanation
Correct option: A
a(t+1t)2a{\left( {t + \frac{1}{t}} \right)^2}
The correct answer is A (a(t+1t)2a{\left( {t + \frac{{1}}{t}} \right)^2}) because it represents the length of the focal chord of the parabola using the parameter t.
17 Medium
17 of 211
Application +4 | -1
Let (x,y)(x, y) be any point on the parabola y2=4xy^2=4 x. Let PP be the point that divides the line segment from (0,0)(0,0) to (x,y)(x, y) in the ratio 2:32: 3. Then locus of PP is
View answer and explanation
Correct option: C
5y2=8x5 y^2=8 x
Option C (5y2=8x5 y^2=8 x) is correct because the locus of the point P that divides the line segment from (0,0) to (x,y) in the ratio 2:3 is 5y2=8x5y^2 = 8x.
18 Medium
18 of 211
Application +4 | -1
If the chord of contact of tangents drawn from PP to the parabola y2=4ax{y^2} = 4ax touches the rectangular hyperbola x2−y2=a2,{x^2} - {y^2} = {a^2}, then PP lies on
View answer and explanation
Correct option: C
4x2+y2=4a24{x^2} + {y^2} = 4{a^2}
Option D (4x2+y2=4a24{x^2} + {y^2} = 4{a^2}) is the correct locus of point P because the chord of contact touches the rectangular hyperbola under the given conditions.
19 Medium
19 of 211
Application +4 | -1
Tangents drawn from the point (c,d)\left( {c,d} \right) to the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 make angles α\alpha and β\beta with the xx-axis. If tan⁡αtan⁡β=1,\tan \alpha \tan \beta = 1, then c2−d2={c^2} - {d^2} =
View answer and explanation
Correct option: C
a2+b2{a^2} + {b^2}
Option C (a2+b2{a^2} + {b^2}) is correct because the condition tan⁡αtan⁡β=1\tan \alpha \tan \beta = 1 leads to the relationship c2−d2=a2+b2{c^2} - {d^2} = {a^2} + {b^2}.
20 Medium
20 of 211
Application +4 | -1
From the point (−1,2)(-1,2), the tangent lines are drawn to the parabola y2=4xy^2=4 x, then the equation of chord of contact is
View answer and explanation
Correct option: B
y=x−1y=x-1
Option B (y=x−1y=x-1) is correct because the equation of the chord of contact is found using the formula T=0T=0.
21 Medium
21 of 211
Application +4 | -1
Let AA be the vertex and LL the length of the latus rectum of the parabola y2−2y−4x−7=0y^2-2 y-4 x-7=0. The equation of a parabola with AA as vertex, 2 L2 \mathrm{~L} the length of the latus rectum and the axis at right angles to that of the given curve is -
View answer and explanation
Correct option: A
x2+4x+8y−4=0x^2+4 x+8 y-4=0
Option A (x2+4x+8y−4=0x^2+4 x+8 y-4=0) is correct because it represents a parabola with the given vertex, latus rectum length, and axis perpendicular to the original parabola.
22 Medium
22 of 211
Application +4 | -1
The equation of the latus rectum of the parabola represented by equation y2+2Ax+2By+C=0y^2+2 A x+2 B y+C=0 is
View answer and explanation
Correct option: C
x=B2−A2−C2Ax=\frac{B^2-A^2-C}{2 A}
Option C (x=B2−A2−C2Ax=\frac{{B^2-A^2-C}}{{2 A}}) is correct because it represents the x-coordinate of the latus rectum, which passes through the focus of the parabola.
23 Medium
23 of 211
Analyzing +4 | -1
Chords of the parabola y2=4xy^2=4 x touch the hyperbola x2−y2=1x^2-y^2=1. The locus of the point of intersection of the tangents drawn to the parabola at the extremities of such chords is
View answer and explanation
Correct option: C
An ellipse
Option C (An ellipse) is correct because the locus of the point of intersection of tangents is an ellipse.
24 Medium
24 of 211
Application +4 | -1
P1:y2=49xP_1: y^2=49 x and P2:x2=4ayP_2: x^2=4 a y are two parabolas. Equation of a tangent to the parabola P1P_1 at a point where it intersects the parabola P2P_2 is:
View answer and explanation
Correct option: C
x−2y+4a=0x-2 y+4 a=0
Option A (x−2y+4a=0x-2 y+4 a=0) is correct because it represents the tangent to the parabola P1P_1 at the point where it intersects P2P_2, derived from solving the system of equations and finding the tangent.
25 Medium
25 of 211
Application +4 | -1
The locus of a foot of perpendicular drawn to the tangent of parabola y2=4axy^2=4 a x from focus, is
View answer and explanation
Correct option: A
x=0x=0
Option B (y=0y=0) is incorrect. Option A (x=0x=0) is correct because the locus of the foot of the perpendicular drawn to the tangent of the parabola y2=4axy^2=4 a x from the focus is the tangent at the vertex, which is the y-axis.
26 Medium
26 of 211
Application +4 | -1
If two normals to the parabola (y−2)2=−12(x+3)(y-2)^2=-12(x+3) intersect at right angles then the chord joining their feet passes through a fixed point whose co-ordinates are:
View answer and explanation
Correct option: C
(−6,2)(-6,2)
Option C ((-6,2)) is correct because the chord joining the feet of two perpendicular normals to a parabola passes through the focus, and the focus of the given parabola is (-6, 2).
27 Medium
27 of 211
Application +4 | -1
The equation of the parabola with focus (a,b)(a, b) and directrix xa+yb=1\frac{x}{a}+\frac{y}{b}=1 is given by
View answer and explanation
Correct option: A
(ax−by)2−2a3x−2b3y+a4+a2b2+b4=0(a x-b y)^2-2 a^3 x-2 b^3 y+a^4+a^2 b^2+b^4=0
Option A ((ax−by)2−2a3x−2b3y+a4+a2b2+b4=0(a x-b y)^2-2 a^3 x-2 b^3 y+a^4+a^2 b^2+b^4=0) is correct because it represents the equation of the parabola with the given focus and directrix, derived using the distance formula.
28 Medium
28 of 211
Application +4 | -1
The eccentricity of the conic represented by the parametric equation x=ρ−t+1,y=ρ+t+1x=\rho-t+1, y=\rho+t+1
View answer and explanation
Correct option: A
1
Option A is correct because the eccentricity of the conic represented by the parametric equations x = t2 - t + 1 and y = t2 + t + 1 is 1, indicating a parabola.
29 Medium
29 of 211
Application +4 | -1
If the equation of normal to the parabola y2=4axy^2=4 a x having slope -2 is given by y+2x+ka=0y+2 x+k a=0, then value of kk.
View answer and explanation
Correct option: D
-12
Option B (12) is correct because the length AA' is calculated using the properties of the normal at the ends of the latus rectum of the parabola.
30 Medium
30 of 211
Application +4 | -1
If x+y=kx+y=k is a normal to the parabola y2=12xy^2=12 x, then it touches the parabola
View answer and explanation
Correct option: A
y2=−36xy^2=-36 x
Option A (y2=−36xy^2=-36 x) is correct because it represents the parabola that the normal line touches, derived from the normal condition and the tangent equation.
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  • ✓ This set contains 211 single-correct-answer MCQs on PARABOLA.
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  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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