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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 61-90 of 586 questions

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61 Medium
61 of 586
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.
62 Easy
62 of 586
Analyzing +4 | -1
If the line ax+y=cax + y = c, touches both the curves x2+y2=1{x^2} + {y^2} = 1 and y2=42x{y^2} = 4\sqrt 2 x, then ∣c∣\left| c \right| is equal to
View answer and explanation
Correct option: C
2\sqrt 2
Option C (2\sqrt 2) is correct because it is found by using the tangent condition for both the circle and the parabola and solving for |c|.
63 Easy
63 of 586
Application +4 | -1
Two common tangents to the circle x2+y2=2a2x^2+y^2=2 a^2 and parabola y2=8axy^2=8ax are
View answer and explanation
Correct option: B
y=±(x+2a)y=\pm(x+2 a)
Option B (y=±(x+2a)y=\pm(x+2 a)) is correct because by finding the tangent to the parabola and then applying the condition for tangency to the circle, we arrive at this equation.
64 Easy
64 of 586
Application +4 | -1
The mid-point of the line joining the common points of the line 2x−3y+8=02x - 3y + 8 = 0 and y2=8x,{y^2} = 8x, is
View answer and explanation
Correct option: B
(5,6)\left( {5,6} \right)
Option B ((5,6)) is correct because it is found by using the T=S' method to find the midpoint of the chord intercepted by the line and parabola.
65 Easy
65 of 586
Analyzing +4 | -1
The normal chord of a parabola y2=4ax{y^2} = 4ax at (x1,x1)\left( {{x_1},{x_1}} \right) subtends a right angle at the
View answer and explanation
Correct option: A
Focus
Option B (A parabola with vertex at (2,1)\left( {2,1} \right)) is correct. By eliminating the parameter 't', we obtain the equation of a parabola with vertex at (2,1).
66 Easy
66 of 586
Application +4 | -1
From an external point PP, pair of tangent lines are drawn to the parabola, y2=4xy^2=4 x. If θ1&θ2\theta_1 \& \theta_2 are the inclinations of these tangents with the axis of xx such that, θ1+θ2=π4\theta_1+\theta_2=\frac{\pi}{4}, then the locus of PP is :
View answer and explanation
Correct option: C
x−y−1=0x-y-1=0
Option C (x−y−1=0x-y-1=0) is correct because it represents the locus of point P derived from the condition on the inclinations of the tangents to the parabola.
67 Medium
67 of 586
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.
68 Medium
68 of 586
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.
69 Easy
69 of 586
Application +4 | -1
The line lx+my+n=0l x+m y+n=0 is a normal to the parabola y2=4axy^2=4 a x, if
View answer and explanation
Correct option: A
al(l2+2m2)+m2n=0a l\left(l^2+2 m^2\right)+m^2 n=0
Option A (al(l2+2m2)+m2n=0a l(l^2+2 m^2)+m^2 n=0) is correct because it represents the condition for the line lx+my+n=0l x+m y+n=0 to be a normal to the parabola y2=4axy^2=4 a x.
70 Easy
70 of 586
Remembering +4 | -1
For the parabola y2=8xy^2=8 x, the length of the latus-rectum is
View answer and explanation
Correct option: A
8
Option A (8) is correct because for the parabola y2=4axy^2 = 4ax, the length of the latus rectum is 4a4a, and in this case, 4a=84a = 8.
71 Easy
71 of 586
Application +4 | -1
If bb and kk are segments of a focal chord of the parabola y2=4axy^2=4 a x, then k=k=
View answer and explanation
Correct option: A
abb−a\frac{a b}{b-a}
Option A (abb−a\frac{a b}{b-a}) is correct because the semi-latus rectum of a parabola is the harmonic mean of the segments of a focal chord, and solving for k gives the correct expression.
72 Easy
72 of 586
Analyzing +4 | -1
The focal chords of the parabola y2=16xy^2 = 16x which are tangent to the circle of radius rr and
centre (6,0) are perpendicular, then the radius rr of the circle is
View answer and explanation
Correct option: B
2\sqrt{2} Units
Option A (944\frac{9}{44}) is correct because after taking the limit and simplifying the summation, the result is 944\frac{9}{44}.
73 Medium
73 of 586
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).
74 Medium
74 of 586
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.
75 Easy
75 of 586
Application +4 | -1
The equation of the parabola whose focus is (3,−4)\left( {3, - 4} \right) and directrix 6x−7y+5=06x - 7y + 5 = 0, is
View answer and explanation
Correct option: A
(7x+6y)2−570x+750y+2100=0{\left( {7x + 6y} \right)^2} - 570x + 750y + 2100 = 0
The correct answer is A ((7x+6y)2−570x+750y+2100=0{\left( {7x + 6y} \right)^2} - 570x + 750y + 2100 = 0) because it is derived by applying the definition of a parabola: the distance to the focus equals the distance to the directrix.
76 Easy
76 of 586
Application +4 | -1
The vertex of the parabola 3x−2y2−4y+7=03 x-2 y^2-4 y+7=0 is
View answer and explanation
Correct option: B
(−3,−1)(-3,-1)
Option C ((-3, -1)) is correct because the vertex of the parabola is (-3, -1) after rewriting the equation in standard form.
77 Easy
77 of 586
Application +4 | -1
The equation of the tangent at a point P(t)P(t) where tt is any parameter to the parabola y2=4axy^2=4 a x, is
View answer and explanation
Correct option: A
yt=x+at2y t=x+a t^2
Option A (yt=x+at2y t=x+a t^2) is correct because it represents the equation of the tangent at a point P(t)P(t) to the parabola y2=4axy^2=4 a x.
78 Easy
78 of 586
Analyzing +4 | -1
The curve described parametrically by the equations x=t2+t+1,y=t2−t+1x=t^2+t+1, y=t^2-t+1, is
View answer and explanation
Correct option: B
A parabola
The correct answer is B (A parabola) because the parametric equations can be reduced to a quadratic equation representing a parabola.
79 Easy
79 of 586
Application +4 | -1
A parabola passing through the point (−4,−2)(-4,-2) has its vertex at the origin and yy-axis as its axis. The latus rectum of the parabola is
View answer and explanation
Correct option: B
8
Option B (8) is correct because substituting the given point into the parabola equation and calculating the latus rectum yields 8.
80 Medium
80 of 586
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.
81 Easy
81 of 586
Application +4 | -1
The equation of the tangents to the parabola y2=9xy^2=9 x which go through the point (4,10)(4,10) are
View answer and explanation
Correct option: D
4y−x=36,4y−9x=44 y-x=36,4 y-9 x=4
Option D (4y−x=36,4y−9x=44 y-x=36,4 y-9 x=4) is correct because these are the equations of the tangents to the parabola that pass through the given point.
82 Easy
82 of 586
Application +4 | -1
The length of the latusrectum of the parabola x2−4x−8y+12=0x^2-4 x-8 y+12=0, is
View answer and explanation
Correct option: C
8
Option C (8) is correct because the length of the latus rectum is the coefficient of the y-term after completing the square.
83 Easy
83 of 586
Application +4 | -1
Locus of the poles of focal chords of a parabola is
View answer and explanation
Correct option: D
The directrix
Option D (The directrix) is correct because the locus of the poles of focal chords of a parabola is the directrix.
84 Medium
84 of 586
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.
85 Easy
85 of 586
Analyzing +4 | -1
The locus of the mid-point of the line segment joining a point on the parabola y2=4axy^2=4 a x and the point of contact of circle drawn on focal distance of the point as diameter with the tangent at the vertex, is
View answer and explanation
Correct option: D
2y2=9ax2 y^2=9 a x
Option D (2y2=9ax2 y^2=9 a x) is correct because the locus of the midpoint is the parabola 2y2=9ax2y^2 = 9ax.
86 Easy
86 of 586
Application +4 | -1
A line LL passing through the focus of the parabola (y−2)2=4(x+1){\left( {y - 2} \right)^2} = 4\left( {x + 1} \right) intersects the parabola in two distinct points. If m be the slope of the line LL, then
View answer and explanation
Correct option: C
m∈(−∞,0)∪(0,∞)m \in \left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)
Option C (m∈(−∞,0)∪(0,∞)m \in ( - \infty ,0) \cup (0,\infty )) is correct because any line passing through the focus of a parabola intersects it at two distinct points, except for the line parallel to the axis (which has a slope of 0).
87 Easy
87 of 586
Application +4 | -1
The equation of a tangent to the hyperbola 4x2−5y2=204{x^2} - 5{y^2} = 20 parallel to the line x−y=2x - y = 2 is
View answer and explanation
Correct option: C
x−y+1=0x - y + 1 = 0
Option C (x−y+1=0x - y + 1 = 0) is correct because it represents a tangent to the hyperbola with a slope of 1, derived using the tangent equation formula.
88 Easy
88 of 586
Application +4 | -1
The vertex of the parabola y2+6x−2y+13=0,{y^2} + 6x - 2y + 13 = 0, is
View answer and explanation
Correct option: B
(−2,1)\left( { - 2,1} \right)
Option B ((−2,1-2, 1)) is correct because it represents the vertex of the given parabola after rewriting the equation in standard form.
89 Medium
89 of 586
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.
90 Medium
90 of 586
Analyzing +4 | -1
The possible coordinate of the point on y2=8xy^2=8 x, which is closest from x2+(y+6)2=1\mathrm{x}^2+(\mathrm{y}+6)^2=1 is :
View answer and explanation
Correct option: A
(2,−4)(2,-4)
Option A ((2,-4)) is correct because it represents the point on the parabola closest to the given circle, found by minimizing the distance between the point on the parabola and the center of the circle.
Revision notes and key takeaways

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