Skip to main content
Back to PARABOLA

Your Progress

0 / 586 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 121-150 of 586 questions

Preparing practice and saved progress…

121 Easy
121 of 586
Analyzing +4 | -1
Let x=2t,  y=t23x = 2t,\,\,y = \frac{{{t^2}}}{3} be a conic. Let SS be the focus and BB be the point on the axis of the conic such that SA⊥BASA \bot BA where AA is any point on the conic. If kk is the ordinate of the centroid of the ΔSAB,\Delta SAB, then lim⁡t→1k \lim\limits_{t \to 1} k\, is equal to
View answer and explanation
Correct option: D
1318\frac{{13}}{{18}}
The correct answer is D (1318\frac{{13}}{{18}}) because it is the limit of the ordinate of the centroid of the triangle SAB as t approaches 1, calculated using the properties of the conic and the given geometric conditions.
122 Medium
122 of 586
Application +4 | -1
Find length of focal chord of parabola y2=8xy^2=8 x which is perpendicular to line x+y=1x+y=1
View answer and explanation
Correct option: B
828 \sqrt{2}
Option B (828 \sqrt{2}) is correct because the length of the focal chord perpendicular to the given line is calculated using the formula 4acsc⁡2(θ)4a \csc^2(\theta), where θ\theta is the angle the chord makes with the x-axis.
123 Medium
123 of 586
Application +4 | -1
The focus of the parabola (y−2)2=20(x+3)(y-2)^2=20(x+3) is
View answer and explanation
Correct option: C
(2,2)(2,2)
Option C (2,2) is correct because the focus of the parabola is found by comparing the given equation with the standard equation and calculating the focus using the vertex and the value of 'a'.
124 Easy
124 of 586
Analyzing +4 | -1
The number of distinct normals that can be drawn from (−2,1)(-2,1) to the parabola y2−4x−2y−3=0y^2-4 x-2 y-3=0, is
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one normal can be drawn from the point (-2, 1) to the given parabola.
125 Medium
125 of 586
Application +4 | -1
If x2+6x+20y−51=0x^2+6 x+20 y-51=0, then the axis of parabola is
View answer and explanation
Correct option: A
x+3=0x+3=0
Option A (x+3=0x+3=0) is correct because the axis of the parabola is found by completing the square and identifying the x-coordinate of the vertex.
126 Easy
126 of 586
Application +4 | -1
The centre of the conic 14x2−4xy+11y2−44x−58y+71=0,14 x^2-4 x y+11 y^2-44 x-58 y+71=0, is
View answer and explanation
Correct option: A
(2,3)(2,3)
Option A (2,3) is correct because it is found by solving the partial derivatives of the conic equation with respect to x and y.
127 Medium
127 of 586
Application +4 | -1
If the line y=mx+cy=m x+c is a tangent to the parabola y2=4a(x+a)y^2=4 a(x+a) then ma+amm a+\frac{a}{m} is equal to
View answer and explanation
Correct option: A
cc
Option A (cc) is correct because if the line y=mx+cy=m x+c is a tangent to the parabola y2=4a(x+a)y^2=4 a(x+a), then ma+amm a+\frac{a}{m} is equal to cc.
128 Medium
128 of 586
Application +4 | -1
The equation of a line touching both the parabolas y2=4xy^2=4 x and x2=−32yx^2=-32 y, is
View answer and explanation
Correct option: D
x−2y+4=0x-2 y+4=0
Option D (x−2y+4=0x-2 y+4=0) is correct because by finding the tangent to the first parabola and then applying the condition for tangency to the second parabola, we arrive at this equation.
129 Easy
129 of 586
Application +4 | -1
The coordinates of the extremities of the latus rectum of the parabola 5y2=4x5 y^2=4 x are
View answer and explanation
Correct option: B
(1/5,2/5),(1/5,−2/5)(1 / 5,2 / 5),(1 / 5,-2 / 5)
Option B ((1/5,2/5),(1/5,−2/5)(1 / 5,2 / 5),(1 / 5,-2 / 5)) is correct because these are the coordinates of the endpoints of the latus rectum for the given parabola.
130 Easy
130 of 586
Application +4 | -1
If the common tangent to the parabolas y2=4xy^2=4 x and x2=4yx^2=4 y also touches the circle x2+y2=c2x^2+y^2=c^2, then cc is equal to
View answer and explanation
Correct option: A
12\frac{1}{\sqrt{2}}
Option A (12\frac{1}{\sqrt{2}}) is correct because the common tangent's distance from the origin equals the radius of the circle, leading to c = 12\frac{1}{\sqrt{2}}.
131 Easy
131 of 586
Application +4 | -1
The area of the triangle inscribed in the parabola y2=4x{y^2} = 4x the ordinates of whose vertices are 1,2 and 4 is
View answer and explanation
Correct option: D
34\frac{3}{4} sq. units
Option D (34\frac{3}{4} sq. units) is correct because it is the area of the triangle calculated using the formula for the area of a triangle inscribed in a parabola.
132 Easy
132 of 586
Application +4 | -1
The equation of the parabola whose vertex and focus are on the positive side of the xx-axis at distances aa and bb respectively from the origin, is
View answer and explanation
Correct option: A
y2=4(b−a)(x−a)y^2=4(b-a)(x-a)
Option A (y2=4(b−a)(x−a)y^2=4(b-a)(x-a)) is correct because it represents a parabola with the given vertex and focus, with the correct latus rectum length.
133 Easy
133 of 586
Application +4 | -1
If (2,0)(2,0) is the vertex and yy-axis the directrix of a parabola, then its focus is
View answer and explanation
Correct option: C
(4,0)(4,0)
Option C ((4,0)(4,0)) is correct because the focus is equidistant from the vertex as the directrix, but on the opposite side.
134 Easy
134 of 586
Application +4 | -1
PQPQ is any focal chord of the parabola y2=32xy^2=32 x. The length of PQP Q can never be less than
View answer and explanation
Correct option: C
32 units
Option C (32 units) is correct because the shortest focal chord is the latus rectum, and for the parabola y2=4axy^2 = 4ax, the length of the latus rectum is 4a4a.
135 Medium
135 of 586
Application +4 | -1
A triangle ABCABC of area Δ\Delta is inscribed in the parabola y2=4ax{y^2} = 4ax such that AA is the vertex and BCBC is a focal chord of the parabola. The difference of the ordinates of BB and CC is
View answer and explanation
Correct option: A
2Δa\frac{{2\Delta }}{a}
The correct answer is 2Δa\frac{{2\Delta }}{a} because it represents the difference of the ordinates of B and C, calculated using the area of the triangle and the properties of the parabola.
136 Medium
136 of 586
Application +4 | -1
The point on y2=4axy^2=4 a x nearest to the focus has its abscissa equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because the distance from any point on the parabola to the focus is minimized when the point is at the vertex.
137 Easy
137 of 586
Application +4 | -1
The equation of the parabola whose vertex is (−1,−2)(-1,-2), axis is vertical, and which passes through the point (3,6)(3,6), is
View answer and explanation
Correct option: A
x2+2x−2y−3=0x^2+2 x-2 y-3=0
Option A (x2+2x−2y−3=0x^2+2 x-2 y-3=0) is correct because substituting the vertex and the given point into the parabola equation yields this result.
138 Easy
138 of 586
Application +4 | -1
If the line x+y−1=0x + y - 1 = 0 is a tangent to the parabola y2−y+x=0{y^2} - y + x = 0, then the point of contact is
View answer and explanation
Correct option: A
(0,1)\left( {0,1} \right)
Option A ((0,1)) is correct because it's the only point that satisfies both the line and parabola equations, indicating it's the point of contact.
139 Easy
139 of 586
Application +4 | -1
The equation of a common tangent to y2=4xy^2=4 x and the curve x2+4y2=8x^2+4 y^2=8 can be
View answer and explanation
Correct option: B
x+2y+4=0x+2 y+4=0
Option B (x+2y+4=0x+2 y+4=0) is correct because it is a common tangent to both the parabola and the ellipse, found by comparing the tangent equations of both curves.
140 Easy
140 of 586
Application +4 | -1
Two perpendicular tangents PA{PA} and PB{PB} are drawn to y2=4ax{y}^2=4 {ax}, minimum length of AB{AB} is equal to
View answer and explanation
Correct option: B
4a4 a
Option B (4 a) is correct because the chord of contact of mutually perpendicular tangents to a parabola is a focal chord, and the minimum length of a focal chord is the latus rectum, which is 4a.
141 Easy
141 of 586
Application +4 | -1
The ends of a line segment are P(1,  3)P\left( {1,\;3} \right) and Q(1,  1).Q\left( {1,\;1} \right). RR is a point on the line segment PQPQ such that PR:QR=1:λPR:QR = 1:\lambda. If RR is an interior point of the parabola y2=4x{y^2} = 4x, then
View answer and explanation
Correct option: A
λ∈(0,  1)\lambda \in \left( {0,\;1} \right)
Option A (λ∈(0,  1)\lambda \in \left( {0,\;1} \right)) is correct because the point R lies inside the parabola only when the ratio λ\lambda is between 0 and 1.
142 Medium
142 of 586
Application +4 | -1
The equation of a tangent to the parabola y2=4axy^2=4 a x which is making an angle θ\theta with xx-axis is
View answer and explanation
Correct option: C
y=xtan⁡θ+acot⁡θy=x \tan \theta+a \cot \theta
Option C (y=xtan⁡θ+acot⁡θy=x \tan \theta+a \cot \theta) is correct because it represents the equation of a tangent to the parabola y2=4axy^2=4 a x with slope tan⁡θ\tan \theta.
143 Easy
143 of 586
Application +4 | -1
The axis of the parabola 9y2−16x−12y−57=09 y^2-16 x-12 y-57=0, is
View answer and explanation
Correct option: A
3y=23 y=2
Option A (3y=23y=2) is correct because it represents the axis of symmetry for the given parabola after completing the square.
144 Easy
144 of 586
Application +4 | -1
If x=my+cx = my + c is a normal to the parabola x2=4  ay,{x^2} = 4\;ay, then the value of cc is
View answer and explanation
Correct option: A
−2  am−am3- 2\;am - a{m^3}
Option A (−2  am−am3- 2\;am - a{m^3}) is correct because this is the standard form for the constant term in the normal equation to the given parabola.
145 Medium
145 of 586
Application +4 | -1
A line passing through (21,30)(21,30) and normal to the curve y=2xy=2 \sqrt{x}. If m\mathrm{m} is slope of the normal then m+6=m+6=
View answer and explanation
Correct option: A
1
Option A (1) is correct because finding the slope of the normal and adding 6 gives the correct result.
146 Medium
146 of 586
Application +4 | -1
The length of the normal of the parabola y2=4x{y^2} = 4x which subtends a right angle at the vertex is
View answer and explanation
Correct option: A
636\sqrt 3
Option A (636\sqrt 3) is correct because it is the length of the normal calculated using the given conditions and properties of the parabola.
147 Easy
147 of 586
Application +4 | -1
The equation of the common tangent to the parabolas y2=4axy^2=4ax and x2=4byx^2=4 b y, is
View answer and explanation
Correct option: B
a1/3x+b1/3y+(ab)2/3=0a^{1 / 3} x+b^{1 / 3} y+(a b)^{2 / 3}=0
Option A (a1/3x+b1/3y+(ab)2/3=0a^{1 / 3} x+b^{1 / 3} y+(a b)^{2 / 3}=0) is correct because it is derived by equating the tangent equations of the two parabolas and solving for the common tangent.
148 Medium
148 of 586
Application +4 | -1
Set of values of m for which a chord of slope m of the circle x2+y2=4{x^2} + {y^2} = 4 touches the parabola y2=4x,{y^2} = 4x, is
View answer and explanation
Correct option: A
(−∞,−2−12)∪(2−12,∞)\left( { - \infty , - \sqrt {\frac{{\sqrt 2 - 1}}{2}} } \right) \cup \left( {\sqrt {\frac{{\sqrt 2 - 1}}{2},\infty } } \right)
Option A ((−∞,−2−12)∪(2−12,∞)\left( { - \infty , - \sqrt {\frac{{\sqrt 2 - 1}}{2}} } \right) \cup \left( {\sqrt {\frac{{\sqrt 2 - 1}}{2},\infty } } \right)) is correct because it represents the set of values of m for which the chord of the circle touches the parabola.
149 Easy
149 of 586
Application +4 | -1
The line x+y=6x+y=6 is a normal to the parabola y2=8xy^2=8 x at the point.
View answer and explanation
Correct option: C
(2,4)(2,4)
Option C ((2,4)) is correct because the slope of the normal is -1, and substituting this into the parabola's normal equation gives the point (2,4).
150 Easy
150 of 586
Analyzing +4 | -1
Consider the parabola y2=8xy^2=8 x. Let Δ1\Delta_1 be the area of the triangle formed by the end points of its latus rectum and the point P(1/2,2)P(1 / 2,2) on the parabola, and Δ2\Delta_2 be the area of the triangle formed by drawing tangents at PP and at the end-points of the latus rectum. Then, Δ1Δ2\frac{\Delta_1}{\Delta_2}, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the area of a triangle formed by three points on a parabola is twice the area of the triangle formed by tangents at those points.
Revision notes and key takeaways

Review PARABOLA in your textbook or class notes, then use this set to check your understanding.

Showing 121-150 of 586 questions

Your practice

0 / 586 answered · 586 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.