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JEE • MATHEMATICS • PARABOLA

Parabola practice · Easy

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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 181-210 of 356 questions

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181 Easy
181 of 356
Application +4 | -1
Tangents PQP Q and PRP R are drawn to the parabola y2=4xy^2=4 x from the point P(4,5)P(4,5) then the area of the triangle PQRP Q R is
View answer and explanation
Correct option: C
272\frac{27}{2}
Option C (272\frac{27}{2}) is correct because it is the area of the triangle calculated using the formula involving the point P and the parameter 'a'.
182 Easy
182 of 356
Analyzing +4 | -1
Angle between the tangents drawn to y2=4x{y^2} = 4x at the points where it is intersected by the line y=x−1y = x - 1 is equal to
View answer and explanation
Correct option: D
π2\frac{\pi }{2}
Option D (π2\frac{\pi }{2}) is correct because the line intersects the parabola at the focus, making it a focal chord, and tangents at the ends of a focal chord are perpendicular.
183 Easy
183 of 356
Application +4 | -1
The equation of the tangent at the vertex of the parabola x2+4x+2y=0{x^2} + 4x + 2y = 0 is
View answer and explanation
Correct option: C
y=2y = 2
Option C (y=2y = 2) is correct because it represents the tangent at the vertex of the given parabola after rewriting the equation in standard form.
184 Easy
184 of 356
Application +4 | -1
If lx+my+n=0l x+m y+n=0 is tangent to the parabola x2=yx^2=y, then
View answer and explanation
Correct option: D
l2=4mnl^2=4 m n
Option D (l2=4mnl^2=4 m n) is correct because it represents the condition for the line to be tangent to the parabola, derived from setting the discriminant of the resulting quadratic equation to zero.
185 Easy
185 of 356
Application +4 | -1
If the line y=mx+cy=m x+c touches the parabola y2=4a(x+a)y^2=4 a(x+a), then
View answer and explanation
Correct option: B
c=am+amc=a m+\frac{a}{m}
Option B (c=am+amc=a m+\frac{a}{m}) is correct because it represents the condition for tangency of the line to the given parabola.
186 Easy
186 of 356
Application +4 | -1
If bb and cc are the lengths of the segments of any focal chord of a parabola y2=4ax{y^2} = 4ax, then the length of the semilatusrectum is
View answer and explanation
Correct option: D
2bcb+c\frac{{2bc}}{{b + c}}
Option D (2bcb+c\frac{{2bc}}{{b + c}}) is correct because the semi-latus rectum of a parabola is the harmonic mean of the segments of any focal chord.
187 Easy
187 of 356
Application +4 | -1
PQP Q 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: A
32
Option A (32) is correct because the length of the focal chord is a(t+1/t)2a(t + 1/t)^2 and the minimum value of (t+1/t)2(t + 1/t)^2 is 4, hence the minimum length is 32.
188 Easy
188 of 356
Analyzing +4 | -1
Let PQP Q be a focal chord of the parabola y2=4axy^2=4 ax such that tangents at PP and QQ meet at a point on the line y=2x+a,a>0y=2 x+a, a>0. If PQP Q subtends an angle θ\theta at the vertex of y2=4axy^2=4 a x, then tan⁡θ=\tan \theta=
View answer and explanation
Correct option: D
−253-\frac{2 \sqrt{5}}{3}
Option D (−253\frac{-2 \sqrt{5}}{3}) is correct because the tangent of the angle subtended by the focal chord PQ at the vertex is calculated considering the obtuse angle.
189 Easy
189 of 356
Application +4 | -1
The equation of the normal at the point (a4,a)\left(\frac{a}{4}, a\right) to the parabola y2=4axy^2=4 a x is
View answer and explanation
Correct option: B
4x+8y−9a=04 x+8 y-9 a=0
Option B (4x+8y−9a=04 x+8 y-9 a=0) is correct because it represents the normal equation at the given point on the parabola.
190 Easy
190 of 356
Application +4 | -1
The equation of the parabola whose axis is vertical and passes through the points (0,0),(3,0)(0,0),(3,0) and (−1,4)(-1,4) is
View answer and explanation
Correct option: A
x2−3x−y=0x^2-3 x-y=0
Option A (x2−3x−y=0x^2-3 x-y=0) is correct because it is the only equation that satisfies all three given points (0,0), (3,0), and (-1,4).
191 Easy
191 of 356
Application +4 | -1
The point of intersection of the curves whose parametric equations are x=t2+1,y=2tx=t^2+1, y=2 t and x=2 s,y=2sx=2 \mathrm{~s}, y=\frac{2}{s}, is given by
View answer and explanation
Correct option: B
(2,2)(2,2)
Option B ((2,2)) is correct because it is the point that satisfies both parametric equations when converted to Cartesian form.
192 Easy
192 of 356
Application +4 | -1
The tangent to the parabola y2=4x{y^2} = 4x at the point where it intersects the circle x2+y2=5{x^2} + {y^2} = 5 in the first quadrant, passes through the point
View answer and explanation
Correct option: B
(34,74)\left( {\frac{3}{4},\frac{7}{4}} \right)
Option B ((34,74)\left( {\frac{3}{4},\frac{7}{4}} \right)) is correct because the tangent to the parabola at the point of intersection with the circle in the first quadrant passes through this point.
193 Easy
193 of 356
Application +4 | -1
The vertex of a parabola is the point (a,b)(a, b) and Latusrectum is of length l.l. If the axis of the parabola is along the positive direction of yy-axis, then its equation is
View answer and explanation
Correct option: B
(x−a)2=l2(2y−2b)(x-a)^2=\frac{l}{2}(2 y-2 b)
Option B ((x−a)2=l2(2y−2b)(x-a)^2=\frac{l}{2}(2 y-2 b)) is correct because it represents a parabola with vertex (a,b), latus rectum l, and axis along the positive y-axis.
194 Easy
194 of 356
Application +4 | -1
If the focus of a parabola is at (0,−3)(0,-3) and its directrix is y=3y=3, then its equation is
View answer and explanation
Correct option: A
x2=−12yx^2=-12 y
Option A (x2=−12yx^2=-12 y) is correct because it satisfies the definition of a parabola with focus (0,-3) and directrix y=3.
195 Easy
195 of 356
Application +4 | -1
If P(t1),Q(t2)P\left(t_1\right), Q\left(t_2\right) and R(t3)R\left(t_3\right) are three points on the parabola y2=4axy^2=4 a x, then area of ΔPQR=\Delta P Q R=
View answer and explanation
Correct option: B
a2∣(t1−t2)(t2−t3)(t3−t1)∣a^2\left|\left(t_1-t_2\right)\left(t_2-t_3\right)\left(t_3-t_1\right)\right|
Option A (a∣(t1−t2)(t2−t3)(t3−t1)∣a|(t_1-t_2)(t_2-t_3)(t_3-t_1)|) is correct because it represents the area of the triangle formed by three points on the parabola in terms of their parameters.

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196 Easy
196 of 356
Application +4 | -1
The equation 16x2+y2+8xy−74x−78y+212=016 x^2+y^2+8 x y-74 x-78 y+212=0 represents
View answer and explanation
Correct option: B
A parabola
Option B (A parabola) is correct because the discriminant Δ≠0\Delta \neq 0 and h2=abh^2 = ab, which are the conditions for a parabola.
197 Easy
197 of 356
Application +4 | -1
The length of the latusrectum of the parabola whose focus is (3,3)\left( {3,3} \right) and directrix is 3x−4y−2=0,  3x - 4y - 2 = 0,\; is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the length of the latus rectum is twice the distance between the focus and the directrix.
198 Easy
198 of 356
Analyzing +4 | -1
Let AA be a point on the xx-axis. Common tangents are drawn from AA to the curves x2+y2=8x^2+y^2=8 and y2=16xy^2=16 x. If one of these tangents touches the two curves at QQ and RR, then (QR)2(Q R)^2 is equal to
View answer and explanation
Correct option: B
72
Option B is correct because the square of the distance between the points of tangency QQ and RR is 72.
199 Easy
199 of 356
Application +4 | -1
If the tangent to the parabola y2=ax{y^2} = ax makes an angle of 450{45^0} with x−axisx - axis, then the point of contact is
View answer and explanation
Correct option: D
(a4,a2)\left( {\frac{a}{4},\frac{a}{2}} \right)
Option D (a4,a2)\left( {\frac{a}{4},\frac{a}{2}} \right) is correct because it is the point of contact where the tangent to the parabola makes a 45-degree angle with the x-axis.
200 Easy
200 of 356
Application +4 | -1
The straight line x+y=kx + y = k touches the parabola y=x−x2,y = x - {x^2}, if k =
View answer and explanation
Correct option: C
1
Option C (1) is correct because the line touches the parabola when the discriminant of the resulting quadratic equation is zero, leading to k=1.
201 Easy
201 of 356
Analyzing +4 | -1
The number of distinct normal that can be drawn from (11/4,1/4)\left( {11/4,1/4} \right) to the parabola y2=4x,{y^2} = 4x, is
View answer and explanation
Correct option: A
3
Option A (3) is correct because there are three distinct normal lines that can be drawn from the given point to the parabola.
202 Easy
202 of 356
Application +4 | -1
A tangent is drawn to the parabola y2=6x{y^2} = 6x, which is perpendicular to the line 2x+y=1.2x + y = 1. Which of the following points does not lie on it?
View answer and explanation
Correct option: C
(5,4)\left( {5,4} \right)
Option C (5,4) is correct because it does not satisfy the equation of the tangent to the parabola.
203 Easy
203 of 356
Application +4 | -1
The number of distinct normals that can be drawn to parabola y2=16x{y^2} = 16x from the point (2,0)\left( {2,0} \right), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one real value of the slope mm satisfies the condition for the normal to pass through the given point.
204 Easy
204 of 356
Application +4 | -1
If the normal to the parabola y2=4axy^2=4 a x at the point (at 2,2at)\left.{ }^2, 2 a t\right) cuts the parabola again at (aT2,2aT)\left({aT}^2, 2 {aT}\right), then
View answer and explanation
Correct option: D
T2≥8T^2 \geq 8
Option D (T2≥8T^2 \geq 8) is correct because the condition for the normal to cut the parabola again is derived from the quadratic equation having real roots, leading to the inequality T2≥8T^2 \geq 8.
205 Easy
205 of 356
Application +4 | -1
The focus of the parabola 4y2−6x−4y=54 y^2-6 x-4 y=5 is
View answer and explanation
Correct option: B
(−5/8,1/2)(-5 / 8,1 / 2)
Option B ((-5 / 8, 1 / 2)) is correct because the focus of the parabola is (-5/8, 1/2) after rewriting the equation in standard form.
206 Easy
206 of 356
Analyzing +4 | -1
If the normals of the parabola y2=4xy^2=4 x drawn at the end points of its latusrectum are tangents to the circle (x−3)2+(y+2)2=r2(x-3)^2+(y+2)^2=r^2, then the value of r2r^2, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the value of r2r^2 is found to be 2 by substituting the center of the circle into the normal equations and solving.
207 Easy
207 of 356
Application +4 | -1
If two tangents drawn from a point PP to the parabola y2=16(x−3){y^2} = 16(x - 3) are at right angles, then the locus of point PP is
View answer and explanation
Correct option: B
x+1=0x + 1 = 0
Option B (x+1=0x + 1 = 0) is correct because the locus of the point P from which two tangents drawn to the parabola are at right angles is the directrix of the parabola.
208 Easy
208 of 356
Analyzing +4 | -1
The triangle formed by the tangents to the parabola y2=4axy^2=4 a x at the ends of the latusrectum and the double ordinate through the focus, is
View answer and explanation
Correct option: C
right-angled isosceles
Option C (right-angled isosceles) is correct because the tangents at the ends of a focal chord intersect at right angles, and the triangle is isosceles due to symmetry.
209 Easy
209 of 356
Application +4 | -1
Equation of the common tangent to y2=4xy^2=4 x and x2=32yx^2=32 y is
View answer and explanation
Correct option: A
x+2y+4=0x+2 y+4=0
Option A (x+2y+4=0x+2y+4=0) is correct because it represents the common tangent found by solving the equations of tangents to both curves and equating them.
210 Easy
210 of 356
Analyzing +4 | -1
The inclination of the normal with positive direction of xx-axis, drawn at another end of a normal to the parabola y2=4xy^2=4 x is always.
View answer and explanation
Correct option: C
more than 60∘60^{\circ}
Option C (more than 60∘60^{\circ}) is correct because the inclination of the normal is always greater than 60∘60^{\circ} as derived using AM >= GM.
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