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JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

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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 481-510 of 586 questions

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481 Easy
481 of 586
Application +4 | -1
The locus of the point of intersection of two tangents to the parabola y2=4axy^2=4ax which include at an angle of 45∘45^{\circ}, is
View answer and explanation
Correct option: A
y2−4ax=(a+x)2y^2-4 a x=(a+x)^2
Option A (y2−4ax=(a+x)2y^2-4 a x=(a+x)^2) is correct because it is derived by substituting α=45∘\alpha = 45^{\circ} into the general equation for the locus of intersection of tangents at an angle α\alpha.
482 Medium
482 of 586
Application +4 | -1
If a focal chord of y2=16xy^2=16 x is a tangent to (x−6)2+y2=2(x-6)^2+y^2=2, then the values of the slope of this chord are
View answer and explanation
Correct option: A
{−1,1}\{-1,1\}
Option A ({−1,1}\left\{-1,1\right\}) is correct because the tangents to the circle that pass through the focus of the parabola have slopes of ±1\pm 1.
483 Medium
483 of 586
Application +4 | -1
The tangents to the parabola y2=4axy^2=4 a x at P(at12,2at1)P\left(a t_1^2, 2 a t_1\right) and Q(at22,2at2)Q\left(a t_2^2, 2 a t_2\right) intersect on its axis, then
View answer and explanation
Correct option: B
t1=−t2t_1=-t_2
Option B (t1=−t2t_1=-t_2) is correct because the point of intersection of the tangents lies on the x-axis, implying that the y-coordinate of the intersection point is zero.
484 Medium
484 of 586
Remembering +4 | -1
The polar of the focus of the parabola is
View answer and explanation
Correct option: C
Directrix
Option C (Directrix) is correct because the polar of the focus of a parabola is its directrix.
485 Medium
485 of 586
Application +4 | -1
The equation of the parabola with (−3,0)(-3,0) as focus, and x+5=0x+5=0 as directrix, is
View answer and explanation
Correct option: C
y2=4(x+4)y^2=4(x+4)
Option C (y2=4(x+4)y^2=4(x+4)) is correct because it represents a parabola opening to the right with vertex at (-4, 0) and focus at (-3, 0).
486 Hard
486 of 586
Analyzing +4 | -1
The set of points (x,y)(x, y) whose distance from the line y=2x+2y=2 x+2 is the same as the distance from (2,0)(2,0) is a parabola. This parabola is congruent to the parabola in standard from y=kx2y=k x^2 for some kk which is equal to
View answer and explanation
Correct option: D
512\frac{\sqrt{5}}{12}
Option D (512\frac{\sqrt{5}}{12}) is correct because equating the distance from a point to a line and a fixed point results in a parabola, and transforming it to standard form reveals k = 512\frac{\sqrt{5}}{12}.
487 Easy
487 of 586
Application +4 | -1
The value of θ∈(0,π/2)\theta \in(0, \pi / 2) for which the line y=xcos⁡θ+4cos⁡3θ−14cos⁡θ−1y=x \cos \theta+4 \cos ^3 \theta-14 \cos \theta-1 is a normal to the parabola y2=16xy^2=16 x, is
View answer and explanation
Correct option: C
π/9\pi / 9
Option C (π/9\pi / 9) is correct because it is the value of θ\theta that satisfies the condition for the given line to be a normal to the parabola.
488 Easy
488 of 586
Application +4 | -1
Length of the normal chord of the parabola, y2=4xy^2=4 x, which makes an angle of π4\frac{\pi}{4} with the axis of xx is:
View answer and explanation
Correct option: B
828 \sqrt{2}
Option B (828 \sqrt{2}) is correct because it is the calculated length of the normal chord using the given angle and parabola equation.
489 Medium
489 of 586
Application +4 | -1
The equation of the common tangent to the hyperbola 3x2−y2=33{x^2} - {y^2} = 3 and to the parabola y2=8x{y^2} = 8x is
View answer and explanation
Correct option: D
2x+y+1=02x + y + 1 = 0
Option D (2x+y+1=02x + y + 1 = 0) is correct because it represents a common tangent to both the hyperbola and the parabola.
490 Easy
490 of 586
Analyzing +4 | -1
The parabolas : ax2+2bx+cy=0a x^2+2 b x+c y=0 and dx2+2ex+fy=0d x^2+2 e x+f y=0 intersect on the line y=1y=1.

If a,b,c,d,e,fa, b, c, d, e, f are positive real numbers and a,ba, b, cc are in G.P., then
View answer and explanation
Correct option: A
da,eb,fc\frac{d}{a}, \frac{e}{b}, \frac{f}{c} are in A.P.
Option A (da,eb,fc\frac{{d}}{{a}}, \frac{{e}}{{b}}, \frac{{f}}{{c}} are in A.P.) is correct because by finding the intersection point and using the G.P. condition, we can show that the ratios are in arithmetic progression.
491 Medium
491 of 586
Analyzing +4 | -1
If two different tangents of y2=4x{y^2} = 4x are the normals to x2=4by,{x^2} = 4by, then
View answer and explanation
Correct option: B
∣b∣<122\left| b \right| < \frac{1}{{2\sqrt 2 }}
Option B (∣b∣<122\left| b \right| < \frac{1}{{2\sqrt 2 }}) is correct because it represents the condition for two different tangents to the parabola y2=4x{y^2} = 4x to be normals to the parabola x2=4by{x^2} = 4by.
492 Medium
492 of 586
Application +4 | -1
If y=mx+cy=m x+\mathrm{c} is the normal at a point on the parabola y2=8x\mathrm{y}^{2}=8 x whose focal distance is 8 units, then ∣c∣|c| is equal to :
View answer and explanation
Correct option: C
10310 \sqrt{3}
Option C (10310 \sqrt{3}) is correct because it represents the absolute value of c in the normal equation, derived from the given focal distance.
493 Easy
493 of 586
Application +4 | -1
If P(at12,2at1)P\left(a t_1^2, 2 a t_1\right) and Q(at22,2at2)Q\left(a t_2^2, 2 a t_2\right) are two points on the parabola y2=4axy^2=4 a x, then the area of the triangle formed by the tangents at PP and QQ and the chord PQP Q, is
View answer and explanation
Correct option: A
12a2∣t1−t2∣3\frac{1}{2} a^2\left|t_1-t_2\right|^3
Option A (12a2∣t1−t2∣3\frac{1}{2} a^2\left|t_1-t_2\right|^3) is correct because it represents the area of the triangle formed by the tangents and the chord, derived from the determinant formula.
494 Medium
494 of 586
Application +4 | -1
The common tangent to the parabola y2=4ax{y^2} = 4ax and x2=4ay,{x^2} = 4ay, is
View answer and explanation
Correct option: A
x+y+a=0x + y + a = 0
Option A (x+y+a=0x + y + a = 0) is correct because it represents the common tangent to both parabolas, derived by equating the tangent equations and solving for the condition of tangency.
495 Easy
495 of 586
Application +4 | -1
The equation to the line touching both the parabolas y2=4x{y^2} = 4x and x2=−32y,{x^2} = - 32y, is
View answer and explanation
Correct option: D
x−2y+4=0x - 2y + 4 = 0
Option C (x−2y−4=0x - 2y - 4 = 0) is incorrect. Option D (x−2y+4=0x - 2y + 4 = 0) is correct because this line is tangent to both parabolas y2=4x{y^2} = 4x and x2=−32y{x^2} = - 32y.

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496 Medium
496 of 586
Application +4 | -1
The line x−1=0x - 1 = 0 is the directrix of the parabola y2−kx+8=0.{y^2} - kx + 8 = 0. Then one of the value of kk is
View answer and explanation
Correct option: C
4
Option C (4) is correct because substituting the directrix equation into the parabola equation and solving for k yields k=4 or k=-8, and 4 is one of the possible values.
497 Medium
497 of 586
Analyzing +4 | -1
The set of points on the axis of the parabola y2=4x+8y^2=4 x+8 from which 3 normals to the parabola are real and distinct is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the set of points from which 3 normals can be drawn is {(x, 0): x > 0}, which is not among the given options.
498 Easy
498 of 586
Analyzing +4 | -1
If a,b,ca, b, c are distinct positive real numbers such that the parabolas y2=4axy^2=4 a x and y2=4c(x−b)y^2=4 c(x-b) will have a common normal, then
View answer and explanation
Correct option: B
ba−c>2\frac{b}{a-c}>2
Option B (ba−c>2\frac{b}{a-c}>2) is correct because it represents the condition for the existence of a common normal to the two parabolas.
499 Easy
499 of 586
Application +4 | -1
If the line 2x−1=02 x-1=0 is the directrix of the parabola y2−kx+6=0y^2-k x+6=0 then the positive values of kk is
View answer and explanation
Correct option: A
4
Option A (4) is correct because by equating the given directrix with the directrix derived from the parabola equation, we find the positive value of k to be 4.
500 Easy
500 of 586
Application +4 | -1
Equation of the directrix of the parabola whose focus is (0,0)(0,0) and the tangent at the vertex is x−y+1=0x-y+1=0 is
View answer and explanation
Correct option: C
x−y+2=0x-y+2=0
Option C (x−y+2=0x-y+2=0) is correct. The equation of the directrix is found by using the distance between the focus and the directrix, which is twice the distance between the focus and the tangent at the vertex.
501 Easy
501 of 586
Application +4 | -1
The focus of the parabola 4y2+12x−20y4 y^2+12 x-20 y +67=0+67=0 is
View answer and explanation
Correct option: C
(−17/4,5/2)(-17 / 4,5 / 2)
Option C ((-17 / 4,5 / 2)) is correct because it represents the focus of the given parabola, found by completing the square and transforming the equation to standard form.
502 Easy
502 of 586
Application +4 | -1
If x−2y−a=0x - 2y - a = 0 is a chord of y2=4ax,{y^2} = 4ax, then its length is
View answer and explanation
Correct option: C
20 a20\,a
Option C (20 a20\,a) is correct because the given line is a focal chord, and its length can be calculated using the formula 4acsc⁡2θ4a \csc^2 \theta.
503 Easy
503 of 586
Application +4 | -1
The latus rectum of the parabola y2−4y−2x−8=0y^2-4 y-2 x-8=0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the latus rectum of the parabola is 2 after rewriting the equation in standard form.
504 Easy
504 of 586
Application +4 | -1
The normal to the parabola y2=4xy^2=4 x at P(1,2)P(1,2) meets the parabola again in QQ, then the coordinates of QQ are
View answer and explanation
Correct option: B
(9,−6)(9,-6)
Option B ((9,-6)) is correct because using the formula to find the other point of intersection of the normal with the parabola yields (9,-6).
505 Easy
505 of 586
Application +4 | -1
If the axis of a parabola is horizontal and it passes through the points (0,0),(0,−1)(0,0),(0,-1) and (6,1)(6,1), then its equation is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the general form of a parabola with a horizontal axis is x=Ay2+By+Cx = Ay^2 + By + C, and none of the given options fit this form after substituting the points.
506 Easy
506 of 586
Analyzing +4 | -1
The parametric representation (2+t2,2  t+1)\left( {2 + {t^2},2\;t + 1} \right) represents
View answer and explanation
Correct option: B
A parabola with vertex at (2,1)\left( {2,1} \right)
Option D (Finitely may solutions) is correct because by solving the system of equations, we find that there are a finite number of solutions for x, y, and z.
507 Medium
507 of 586
Analyzing +4 | -1
Let a tangent to the curve y2=24xy^2 = 24x meet the curve xy=2xy = 2 at the points AA and B.B. Then the mid points of such line segments ABAB lie on a parabola with the
View answer and explanation
Correct option: A
Directrix 4x=34x = 3
Option A (Directrix 4x=34x = 3) is correct because after finding the locus of the midpoint of the chord, the directrix of the resulting parabola is 4x=34x=3.
508 Easy
508 of 586
Analyzing +4 | -1
If the parabolas y2=4b(x−c){y^2} = 4b(x - c) and y2=8ax {y^2} = 8ax\, have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)?
View answer and explanation
Correct option: C
(1, 1, 3)
Option C (1, 1, 3) is the correct choice because it satisfies the condition for the parabolas to have a common normal other than the axis.
509 Medium
509 of 586
Application +4 | -1
If two normals to a parabola y2=4axy^2=4 a x 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: B
(a,0)({a}, 0)
Option B ((aa, 0)) is correct because the chord joining the feet of two perpendicular normals to the parabola passes through the fixed point (aa, 0).
510 Easy
510 of 586
Analyzing +4 | -1
If parabolas y2=axy^2=a x and 25[(x−3)2+(y+2)2]=(3x−4y−2)225\left[(x-3)^2+(y+2)^2\right]=(3 x-4 y-2)^2 are equal, then the value of aa is
View answer and explanation
Correct option: B
6
Option B (6) is correct because the value of 'a' is determined by equating the latus rectum of both parabolas.
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