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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 421-450 of 586 questions

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421 Easy
421 of 586
Analyzing +4 | -1
If a line along a chord of the circle 4x2+4y2+120x+675=04{x^2} + 4{y^2} + 120x + 675 = 0 passes through the point (-30, 0) and is tangent to the parabola y2=30x{y^2} = 30x, then the length of this chord is
View answer and explanation
Correct option: D
353\sqrt 5
The correct answer is D (353\sqrt 5) because it is the length of the chord calculated using the distance from the center of the circle to the tangent line.
422 Easy
422 of 586
Application +4 | -1
The focus of the parabola x2+2y+6x=0{x^2} + 2y + 6x = 0 is
View answer and explanation
Correct option: A
(−3,4)\left( { - 3,4} \right)
Option A ((-3,4)) is correct because it is found by completing the square and transforming the equation to standard form to identify the focus.
423 Easy
423 of 586
Application +4 | -1
The vertex of the parabola whose focus is (−1,1)(-1,1) and directrix is 4x+3y−24=04 x+3 y-24=0 is
View answer and explanation
Correct option: D
(1,5/2)(1 , 5 / 2)
Option D ((1,5/2)(1 , 5 / 2)) is correct because the vertex is the midpoint between the focus and the foot of the perpendicular from the focus to the directrix.
424 Easy
424 of 586
Application +4 | -1
The point of the contact of the tangent to the parabola y2=4axy^2=4 a x, which makes an angle of 60∘60^{\circ} with xx-axis, is
View answer and explanation
Correct option: A
(a3,2a3)\left(\frac{a}{3}, \frac{2 a}{\sqrt{3}}\right)
Option A ((a3,2a3)\left(\frac{a}{3}, \frac{2 a}{\sqrt{3}}\right)) is correct because it represents the point of contact of the tangent to the parabola y2=4axy^2=4 a x making an angle of 60∘60^{\circ} with the x-axis.
425 Medium
425 of 586
Application +4 | -1
The equation of a parabola which passes through the intersection of a straight line x+y=0x + y = 0 and the circle x2+y2+4y=0{x^2} + {y^2} + 4y = 0 is
View answer and explanation
Correct option: C
y2=2x{y^2} = 2x
Option C (y2=2x{y^2} = 2x) is correct because it passes through the intersection points (0,0) and (2,-2) of the given line and circle.
426 Easy
426 of 586
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 1:3. Then the locus of PP is
View answer and explanation
Correct option: C
y2=xy^2=x
Option C (y2=xy^2=x) is correct because it is the locus of the point P that divides the line segment from (0,0) to (x,y) in the ratio 1:3.
427 Easy
427 of 586
Application +4 | -1
The ends of latus rectum of parabola x2+8y=0x^2+8 y=0 are
View answer and explanation
Correct option: C
(−4,−2)(-4,-2) and (4,−2)(4,-2)
Option C ((−4,−2)(-4,-2) and (4,−2)(4,-2)) is correct because these points satisfy the parabola equation and are at a distance of 2a from the focus.
428 Hard
428 of 586
Application +4 | -1
Radius of circle touching the parabola y2=4xy^2=4 x at point (1,2)(1,2) and passing through its focus is
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because it's the radius of the circle that touches the parabola at (1,2) and passes through the focus (1,0).
429 Easy
429 of 586
Application +4 | -1
If VV and SS are respectively the vertex and focus of the parabola y2+6y+2x+5=0y^2+6 y+2 x+5=0, then SV=S V=
View answer and explanation
Correct option: B
1/21 / 2
Option B (1/21/2) is correct because the distance between the vertex and focus is one-fourth of the latus rectum length.
430 Easy
430 of 586
Application +4 | -1
The length of the latusrectum of the parabola 169{(x−1)2+(y−3)2}=(5x−12y+7)2169\left\{(x-1)^2+(y-3)^2\right\}=(5 x-12 y+7)^2, is
View answer and explanation
Correct option: D
4813\frac{48}{13}
Option D (4813\frac{{48}}{{13}}) is correct because the length of the latus rectum is twice the perpendicular distance from the focus to the directrix.
431 Medium
431 of 586
Application +4 | -1
If the focus of parabola (y−k)2=4(x−h)(y-k)^2=4(x-h) always lies between the line x+y=1x+y=1 and x+y=3\mathrm{x}+\mathrm{y}=3 then -
View answer and explanation
Correct option: A
0<h+k<20<\mathrm{h}+\mathrm{k}<2
Option A (0<h+k<20<\mathrm{h}+\mathrm{k}<2) is correct because the focus of the parabola must lie between the given lines, leading to this inequality.
432 Hard
432 of 586
Analyzing +4 | -1
In a parabola y2=4ax,y^2 = 4ax, two points PP and QQ are taken such that the tangents drawn to parabola at these points
meet at directrix in R.R. Focus of locus of circumcentre of ΔPQR\Delta PQR will be:
View answer and explanation
Correct option: C
(3a2,0)\left( {\frac{{3a}}{2},0} \right)
Option C (3a2,0)\left( \frac{{3a}}{{2}},0 \right) is correct because the locus of the circumcentre is a parabola with focus at (3a2,0)\left( \frac{{3a}}{{2}},0 \right).
433 Easy
433 of 586
Application +4 | -1
If the three normals drawn to the parabola, y2=2x{y^2} = 2x pass through the point (a, 0)\left( {a,\,0} \right) a≠0a \ne 0, then ‘aa’ must be greater than:
View answer and explanation
Correct option: A
1
Option A (1) is correct because the condition for three normals to pass through a point on the axis of the parabola requires 'a' to be greater than 1.
434 Medium
434 of 586
Application +4 | -1
The tangent to the parabola y2=16x{y^2} = 16x, which is perpendicular to a line y−3x−1=0y - 3x - 1 = 0, is
View answer and explanation
Correct option: A
3y+x+36=03y + x + 36 = 0
Option A (3y+x+36=03y + x + 36 = 0) is correct because it represents the tangent to the parabola with a slope that is the negative reciprocal of the given line's slope.
435 Easy
435 of 586
Application +4 | -1
The vertex of the parabola x2+8x+12y+4=0x^2+8 x+12 y+4=0, is
View answer and explanation
Correct option: A
(−4,1)(-4,1)
Option A ((-4,1)) is correct because completing the square and rewriting the equation reveals the vertex form of the parabola.

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436 Easy
436 of 586
Application +4 | -1
The normal at the point (bt12,2bt1)\left(b t_1^2, 2 b t_1\right) on a parabola meets the parabola again in the point (bt22,2bt2)\left(b t_2^2, 2 b t_2\right), then
View answer and explanation
Correct option: B
t2=−t1−2t1t_2=-t_1-\frac{2}{t_1}
Option B (t2=−t1−2t1t_2=-t_1-\frac{2}{t_1}) is correct because it represents the relationship between the parameters of the points where the normal intersects the parabola.
437 Easy
437 of 586
Application +4 | -1
The equation of 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 the equation of a parabola with vertex at (a,0) and focus at (a1,0).
438 Medium
438 of 586
Application +4 | -1
A ray of light moving parallel to the xx-axis gets reflected from parabolic mirror whose equation is (y−3)2=8(x+2)(y-3)^2=8(x+2). After reflection, the ray must pass through
View answer and explanation
Correct option: A
(0,3)(0,3)
Option A ((0,3)) is correct because a ray parallel to the x-axis, after reflection from a parabolic mirror, passes through the focus of the parabola.
439 Easy
439 of 586
Application +4 | -1
If PSQ is a focal chord of the parabola y2=8xy^2=8 x such that SP=6S P=6, then SQ=S Q=
View answer and explanation
Correct option: C
3
Option C (3) is correct because the semi-latus rectum is the harmonic mean of SP and SQ, and solving for SQ gives the correct value.
440 Easy
440 of 586
Application +4 | -1
Tangent to the parabola y=x2+6y=x^2+6 at (1,7)(1,7) touches the circle x2+y2+16x+12y+c=0x^2+y^2+16 x+12 y+c=0 at the point
View answer and explanation
Correct option: C
(−6,−7)(-6,-7)
Option C ((-6,-7)) is correct because it is the point of tangency between the line and the circle.
441 Easy
441 of 586
Application +4 | -1
If y=2x+ky = 2x + k is a tangent to the curve x2=4y,{x^2} = 4y, then kk is equal to
View answer and explanation
Correct option: C
−4- 4
Option C (-4) is correct because substituting the tangent equation into the parabola equation and solving for the tangency condition yields k = -4.
442 Medium
442 of 586
Application +4 | -1
The mid-point of the chord intercepted on the line 4x−3y+4=04 x-3 y+4=0 by the parabola y2=8xy^2=8 x, is
View answer and explanation
Correct option: A
(5/4,3)(5 / 4,3)
Option A ((5/4,3)(5 / 4,3)) is correct because the midpoint is found by equating the equation of the chord with the given line and solving for the coordinates of the midpoint.
443 Medium
443 of 586
Application +4 | -1
The angle between the tangents drawn from the point (3,4)\left( {3,4} \right) to the parabola y2−2y+4x=0{y^2} - 2y + 4x = 0 is
View answer and explanation
Correct option: A
tan⁡−1(85/7){\tan ^{ - 1}}\left( {8\sqrt 5 /7} \right)
Option A (tan⁡−1(85/7){\tan ^{ - 1}}\left( {8\sqrt 5 /7} \right)) is correct because it is the angle between the tangents calculated using the properties of the parabola and the tangent equation.
444 Easy
444 of 586
Application +4 | -1
If the line x+y−1=0x+y-1=0 touches the parabola y2=kxy^2=k x, then the value of kk, is
View answer and explanation
Correct option: B
-4
Option B (-4) is correct because substituting the line equation into the parabola equation and applying the tangency condition (discriminant equals zero) yields k = -4.
445 Hard
445 of 586
Understanding +4 | -1
The circle on focal radii of a parabola as diameter touches the
View answer and explanation
Correct option: C
Tangent at the vertex
The correct answer is C (Tangent at the vertex) because the circle described on the focal radii of a parabola as diameter touches the tangent at the vertex.
446 Medium
446 of 586
Analyzing +4 | -1
Set of values of ' hh ' for which the number of distinct common normals of (x−2)2=4(y−3)(x-2)^2=4(y-3) and x2+y2−2x−hy−c=0(c>0)x^2+y^2-2 x-h y-c=0(c>0) is 3 , is
View answer and explanation
Correct option: D
(10,∞)(10, \infty)
Option D ((10,∞)(10, \infty)) is correct because the condition for the number of distinct common normals of the given parabola and circle to be 3 leads to this interval for the values of 'h'.
447 Medium
447 of 586
Application +4 | -1
If point (2a, a) lies inside the parabola x2−2x−4y+3=0x^2-2 x-4 y+3=0 then interval of values of aa is
View answer and explanation
Correct option: B
(12,32)\left(\frac{1}{2}, \frac{3}{2}\right)
Option B (12,32)\left(\frac{1}{2}, \frac{3}{2}\right) is correct because substituting the point (2a, a) into the inequality derived from the parabola equation yields the correct interval for a.
448 Easy
448 of 586
Application +4 | -1
If a focal chord of the parabola y2=ax{y^2} = ax is 2x−y−8=0,2x - y - 8 = 0, then the equation of the directrix is
View answer and explanation
Correct option: A
x+4=0x + 4 = 0
Option A (x+4=0x + 4 = 0) is correct because the focal chord passes through the focus, allowing us to determine 'a' and subsequently the directrix.
449 Easy
449 of 586
Application +4 | -1
The length of latus rectum of the parabola 4y2+2x−20y4 y^2+2 x-20 y +17=0+17=0 is
View answer and explanation
Correct option: C
1/21 / 2
Option C (1/21 / 2) is correct because the latus rectum of the parabola is 1/2 after rewriting the equation in standard form.
450 Easy
450 of 586
Analyzing +4 | -1
The chord ABAB of the parabola y2=4ax{y^2} = 4ax cuts the axis of the parabola at CC. If A=(at12,  2at1),  B=(at22,  2at2)A = \left( {at_1^2,\;2a{t_1}} \right),\;B = \left( {at_2^2,\;2a{t_2}} \right) and AC:AB=1:3,AC:AB = 1:3, then
View answer and explanation
Correct option: B
t2+2t1=0{t_2} + 2{t_1} = 0
Option B (t2+2t1=0{t_2} + 2{t_1} = 0) is correct because it satisfies the condition AC:AB = 1:3, where C lies on the x-axis.
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