Practice JEE Mathematics Parabola MCQs Set G provided below. The MCQ Questions for JEE Parabola Mathematics with answers and follow the latest JEE (Main)/ NCERT and KVS patterns. Refer to more Chapter-wise MCQs for JEE (Main) JEE Mathematics and also download more latest study material for all subjects
MCQ for JEE Mathematics Parabola
JEE Mathematics students should review the 50 questions and answers to strengthen understanding of core concepts in Parabola
Parabola MCQ Questions JEE Mathematics with Answers
Question. If the line \( x - 1 = 0 \) is the directrix of the parabola \( y^2 - kx + 8 = 0 \), then one of the values of 'k' is
(A) 1/8
(B) 8
(C) 4
(D) 1/4
Answer: (C) 4
\( y^2 = kx - 8 \)
\( y^2 = k\left(x - \frac{8}{k}\right) \)
\( X - \frac{8}{k} = -\frac{k}{4} \)
\( x = \frac{8}{k} - \frac{k}{4} = 1 \)
\( 32 - k^2 = 4k \)
\( k^2 + 4k - 32 = 0 \)
\( k = -8 \)
\( k = 4 \)
Question. If \( x + y = k \) is normal to \( y^2 = 12x \), then 'k' is
(A) 3
(B) 9
(C) -9
(D) -3
Answer: (B) 9
\( y = -x + k \) ... (1)
\( y = mx - 2am - am^3 \), \( a = 3 \)
\( m = -1 \) from Eq\( ^n \) (1)
\( y = -x + 2a + a \) ... (2)
(1) & (2) are same
\( k = 3a \)
\( k = 9 \) (\( a = 3 \))
Question. The equation of the common tangent touching the circle \( (x - 3)^2 + y^2 = 9 \) and the parabola \( y^2 = 4x \) above the x-axis is
(A) \( \sqrt{3}y = 3x + 1 \)
(B) \( \sqrt{3}y = -(x + 3) \)
(C) \( \sqrt{3}y = x + 3 \)
(D) \( \sqrt{3}y = -(3x + 1) \)
Answer: (C) \( \sqrt{3}y = x + 3 \)
\( y^2 = 4x \)
Tangent to the parabola
\( y = mx + \frac{1}{m} \) ; \( a = 1 \)
\( m^2x - my + 1 = 0 \)
C.O.T for circle \( p = r \)
\( c(3, 0) \), \( r = 3 \)
\( \left| \frac{3m^2 + 1}{\sqrt{m^4 + m^2}} \right| = 3 \)
\( (3m^2 + 1)^2 = 9(m^4 + m^2) \)
\( 9m^4 + 6m^2 + 1 = 9m^4 + 9m^2 \)
\( 3m^2 = 1 \)
\( \implies m = \pm \frac{1}{\sqrt{3}} \)
But above the axis
\( m = \frac{1}{\sqrt{3}} \)
\( y = \frac{x}{\sqrt{3}} + \sqrt{3} \)
\( \sqrt{3}y = x + 3 \)
Question. The equation of the directrix of the parabola, \( y^2 + 4y + 4x + 2 = 0 \)
(A) \( x = -1 \)
(B) \( x = 1 \)
(C) \( x = -3/2 \)
(D) \( x = 3/2 \)
Answer: (D) \( x = 3/2 \)
\( y^2 + 4y + 4 + 4x + 2 - 4 = 0 \)
\( (y + 2)^2 + 4x - 2 = 0 \)
\( (y + 2)^2 = -4x + 2 \)
\( (y + 2)^2 = -4\left(x - \frac{1}{2}\right) \)
\( Y^2 = -4aX \)
\( a = 1 \)
\( X = a \)
\( x - \frac{1}{2} = 1 \)
\( \implies x = \frac{3}{2} \)
Question. The locus of the mid-point of the line segment joining the focus to a moving point on the parabola \( y^2 = 4ax \) is another parabola with directrix
(A) \( x = -a \)
(B) \( x = -a/2 \)
(C) \( x = 0 \)
(D) \( x = a/2 \)
Answer: (C) \( x = 0 \)
\( h = \frac{at^2 + a}{2} \)
\( 2h - a = at^2 \) ... (1)
\( k = \frac{2at + 0}{2} \)
\( t = \frac{k}{a} \) ... (2)
\( 2h - a = a\left(\frac{k^2}{a^2}\right) \)
\( k^2 = a(2h - a) \)
\( k^2 = 2a\left(h - \frac{a}{2}\right) \)
\( y^2 = 2a\left(x - \frac{a}{2}\right) \)
Directrix
\( x - \frac{a}{2} = -\frac{a}{2} \)
\( \implies x = 0 \)
Question. The equation of the common tangent to the curves \( y^2 = 8x \) and \( xy = -1 \) is
(A) \( 3y = 9x + 2 \)
(B) \( y = 2x + 1 \)
(C) \( 2y = x + 8 \)
(D) \( y = x + 2 \)
Answer: (D) \( y = x + 2 \)
\( y^2 = 8x \)
\( xy = -1 \)
tangent to above parabola
\( y = mx + \frac{2}{m} \) ... (1)
\( m^2x - my + 2 = 0 \)
\( m^2x + \frac{m}{x} + 2 = 0 \)
\( m^2x^2 + 2x + m = 0 \)
\( D = 0 \)
\( 4 - 4m^3 = 0 \)
\( m = 1 \)
from Eq\( ^n \) (1)
\( y = x + 2 \)
Question. The slope of the focal chords of the parabola \( y^2 = 16x \) which are tangents to the circle \( (x - 6)^2 + y^2 = 2 \) are
(A) \( \pm 2 \)
(B) \( -1/2, 2 \)
(C) \( \pm 1 \)
(D) \( -2, 1/2 \)
Answer: (C) \( \pm 1 \)
\( y^2 = 16x \) (\( a = 4 \))
chord with two points
\( 2x - (t_1 + t_2)y + 2at_1t_2 = 0 \) ... (1)
Passes through focus (4, 0)
\( 8 + 8t_1t_2 = 0 \)
\( \implies t_1t_2 = -1 \) ... (2)
Eq\( ^n \) (1) is tangent to given circle
\( c(6, 0) \); \( r = \sqrt{2} \)
\( \left| \frac{12 + 2at_1t_2}{\sqrt{4 + (t_1 + t_2)^2}} \right| = \sqrt{2} \)
\( 16 = 2[4 + (t_1 + t_2)^2] \)
\( 4 = (t_1 + t_2)^2 \)
\( t_1 + t_2 = \pm 2 \)
Slope = \( \frac{2}{t_1 + t_2} = \pm 1 \)
Question. The angle between the tangents drawn from the point (1, 4) to the parabola \( y^2 = 4x \) is
(A) \( \pi/2 \)
(B) \( \pi/3 \)
(C) \( \pi/4 \)
(D) \( \pi/6 \)
Answer: (B) \( \pi/3 \)
\( y^2 = 4x \quad a = 1 \)
POI of tangent
\( [at_1t_2, a(t_1 + t_2)] \)
\( t_1t_2 = 1 \)
\( t_1 + t_2 = 4 \)
General tangent \( ty = x + at^2 \)
\( m_1 = \frac{1}{t_1} \)
\( m_2 = \frac{1}{t_2} \)
\( \tan\theta = \left| \frac{\frac{1}{t_1} - \frac{1}{t_2}}{1 + \frac{1}{t_1t_2}} \right| = \left| \frac{t_1 - t_2}{1 + t_1t_2} \right| = \frac{\sqrt{(t_1 + t_2)^2 - 4t_1t_2}}{2} \)
\( = \frac{\sqrt{16 - 4}}{2} \)
\( \tan\theta = \sqrt{3} \)
\( \theta = \frac{\pi}{3} \)
Question. The axis of parabola is along the line \( y = x \) and the distance of vertex from origin is \( \sqrt{2} \) and that of origin from its focus is \( 2\sqrt{2} \). If vertex and focus both lie in the 1\( ^{st} \) quadrant, then the equation of the parabola is
(A) \( (x + y)^2 = (x - y - 2) \)
(B) \( (x - y)^2 = (x + y - 2) \)
(C) \( (x - y)^2 = 4(x + y - 2) \)
(D) \( (x - y)^2 = 8(x + y - 2) \)
Answer: (D) \( (x - y)^2 = 8(x + y - 2) \)
Eq\( ^n \) of directrix \( x + y = 0 \)
Hence Eq\( ^n \) of the parabola is
\( \frac{x + y}{\sqrt{2}} = \sqrt{(x - 2)^2 + (y - 2)^2} \)
Eq\( ^n \) of Parabola
\( (x - y)^2 = 8(x + y - 2) \)
Question. The equations of common tangents ot the parabola \( y = x^2 \) and \( y = -(x - 2)^2 \) is/are
(A) \( y = 4(x - 1) \)
(B) \( y = 0 \)
(C) \( y = -4(x - 1) \)
(D) \( y = -30x - 50 \)
Answer: (A), (B)
Eq\( ^n \) of tangent to \( (x - 2)^2 = -y \) is
\( y = m(x - 2) + \frac{m^2}{4} \) ... (2)
(1) & (2) are same
\( m = 0 \) or \( m = 4 \)
\( y = 0 \) and \( y = 4x - 4 \)
Question. Statement-1: The curve \( y = \frac{-x^2}{2} + x + 1 \) is symmetric with respect to the line \( x = 1 \).
Statement -2: A parabola is symmetric about its axis. [JEE 2007, 4]
(A) Statement-1 is true, statement-2 is true ; statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true ; statement-2 is NOT a correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false
(D) Statement-1 is false, statement-2 is true
Answer: (A) Statement-1 is true, statement-2 is true ; statement-2 is correct explanation for statement-1.
\( y = -\frac{x^2}{2} + x + 1 \)
\( y - \frac{3}{2} = -\frac{1}{2}(x - 1)^2 \)
\( (x - 1)^2 = -2\left(y - \frac{3}{2}\right) \)
symmetric about \( x = 1 \)
Comprehension
Consider the circle \( x^2 + y^2 = 9 \) and the parabola \( y^2 = 8x \). They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. [JEE 2007, 4 + 4 + 4]
Question. The ratio of the areas of the triangles PQS and PQR is
(A) \( 1 : \sqrt{2} \)
(B) \( 1 : 2 \)
(C) \( 1 : 4 \)
(D) \( 1 : 8 \)
Answer: (C) \( 1 : 4 \)
\( P(1, 2\sqrt{2}) \)
\( Q(1, -2\sqrt{2}) \)
Area of \( \Delta PQR = \frac{1}{2}(4\sqrt{2})(8) = 16\sqrt{2} \)
Area of \( \Delta PQS = \frac{1}{2}(4\sqrt{2})(2) = 4\sqrt{2} \)
Ratio of area of \( \Delta PQS : \Delta PQR \) is 1 : 4
Question. The radius of the circumcircle of the triangle PRS is
(A) 5
(B) \( 3\sqrt{3} \)
(C) \( 3\sqrt{2} \)
(D) \( 2\sqrt{3} \)
Answer: (B) \( 3\sqrt{3} \)
Eq\( ^n \) of circumcircle of \( \Delta PRS \)
\( (x + 1) (x - 9) + y^2 + \lambda y = 0 \)
It will passes through \( (1, 2\sqrt{2}) \)
\( \implies \lambda = 2\sqrt{2} \)
Equation of circumcircle
\( x^2 + y^2 - 8x + 2\sqrt{2}y - 9 = 0 \)
Hence radius = \( 3\sqrt{3} \)
Aliter
Let \( \angle PSR = \theta \)
\( \sin\theta = \frac{2\sqrt{2}}{2\sqrt{3}} \)
\( PR = 6\sqrt{2} = 2R\sin\theta \)
\( R = 3\sqrt{3} \)
Question. The radius of the incircle of the triangle PQR is
(A) 4
(B) 3
(C) 8/3
(D) 2
Answer: (D) 2
Radius of incircle in \( r = \frac{\Delta}{S} \)
\( \Delta = 16\sqrt{2} \)
\( S = \frac{6\sqrt{2} + 6\sqrt{2} + 4\sqrt{2}}{2} = 8\sqrt{2} \)
\( r = \frac{16\sqrt{2}}{8\sqrt{2}} = 2 \)
Question. The tangent PT and the normal PN to the parabola \( y^2 = 4ax \) at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose
(A) vertex is \( (2a/3, 0) \)
(B) directrix is \( x = 0 \)
(C) latusrectum is \( 2a/3 \)
(D) focus is \( (a, 0) \)
Answer: (A), (D)
\( G \equiv (h, k) \)
\( h = \frac{2a + at^2}{3} \)
\( k = \frac{2at}{3} \)
\( \frac{3h - 2a}{a} = \frac{9k^2}{4a^2} \)
Required Parabola
\( \frac{9y^2}{4a^2} = \frac{3x - 2a}{a} = \frac{3}{a}\left(x - \frac{2a}{3}\right) \)
\( y^2 = \frac{4a}{3}\left(x - \frac{2a}{3}\right) \)
Vertex : \( \left(\frac{2a}{3}, 0\right) \)
Focus : \( (a, 0) \)
Question. Let A and B be two distinct points on the parabola \( y^2 = 4x \). If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be
(A) \( -1/r \)
(B) \( 1/r \)
(C) \( 2/r \)
(D) \( -2/r \)
Answer: (C), (D)
\( A(t_1^2, 2t_1) \), \( B(t_2^2, 2t_2) \)
Centre : \( \left[ \frac{t_1^2 + t_2^2}{2}, t_1 + t_2 \right] \)
\( t_1 + t_2 = \pm r \)
slope of chord = \( \frac{2}{t_1 + t_2} = \pm \frac{2}{r} \)
Free study material for Conic Sections
MCQs for Parabola Mathematics JEE
Students can use these MCQs for Parabola to quickly test their knowledge of the chapter. These multiple-choice questions have been designed as per the latest syllabus for JEE Mathematics released by JEE (Main). Our expert teachers suggest that you should practice daily and solving these objective questions of Parabola to understand the important concepts and better marks in your school tests.
Parabola NCERT Based Objective Questions
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