Students can find the JEE Advanced Maths parabola previous year questions and solutions on this page. Parabola is an important topic in the conic section. It is easy to score marks from this topic if you are thorough with the concepts. Practising previous years’ questions will help students to find their weaker areas and prepare better for them. Solutions are also available in PDF format for free. Students can easily download and learn the solutions offline, too.
Download Parabola Previous Year Solved Questions PDF
Question 1: Let the curve C be the mirror image of the parabola y2 = 4x with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line y = -5, then the distance between A and B is
Solution:
Consider the image of the line y = −5 in the line x + y + 4 = 0 .
To find the image line put the value of y in the equation x + y+ 4 =0
=> x – 5 + 4 = 0
=> x = 1
The image will be the line x = 1.
Find intersection point of line x = 1 and y2 = 4x
put x = 1 in y2 = 4x.
we get y = 2, -2.
So the line x = 1 will intersect the parabola y2 = 4x at the points (1, −2) and (1, 2).
Let the mirror image of these points in the line x+y+4=0 are points A and B.
The distance between AB and its pre-image is the same.
Find distance between (1,−2) and (1,2) using the distance formula.
AB = √(42 + 0) = 4
Distance between A and B is 4 units.
Question 2: If the normals of the parabola y2 = 4x drawn at the endpoints of its latus rectum are tangents to the circle (x-3)2 + (y-2)2 = r2, then the value of r2 is
Solution:
Endpoints of latus rectum of y2 = 4x are (1, ±2)
Equation of normal to y2 = 4x at (1, 2) is
(y-2) = -(x-1)
=> x+y-3 = 0
As it is tangent to circle (x-3)2 + (y+2)2 = r2
So |(3+-2-3)/√2 | = r
=> r2 = 2
Question 3: The point of intersection of the tangents at the ends of the latus rectum of the parabola y2 = 4x is
Solution:
Given parabola is y2 = 4x
=> a = 1
Endpoints of latus rectum are (1, 2) and (1, -2)
The equation of tangent to parabola at any point (x1, y1) is
yy1 = 2a(x+x1)) ..(i)
The equation of the tangent to the end L(1, 2) of the latus rectum is
2y = 2(x+1)
=> x-y+1 = 0 ..(ii)
The equation of the tangent to the end L(1, -2) of the latus rectum is
-2y = 2(x+1)
=> x+y+1 = 0 ..(iii)
Solving (ii) and (iii), we get
x = -1 and y = 0
Hence the point of intersection is (-1, 0).
Question 4: If a chord, which is not a tangent, of the parabola y2 = 16x has the equation 2x + y = p, and midpoint (h, k), then which of the following is(are) possible value(s) of p, h, and k?
(a) p = -2, h = 2, k = -4
(b) p = -1, h = 1, k = -3
(c) p = 2, h = 3, k = -4
(d) p = 5, h = 4, k = -3
Solution:
If (h, k) is the midpoint of chord of parabola y2 = 16x, then the equation of the chord is
T = S1
=> yk – 8(x+h) = k2-16h
=> 8x-ky = 8h – k2 ..(i)
Equation of chord is 2x+y = p ..(ii)
(i) and (ii) are identical lines.
So 8/2 = -k/1 = (8h-k2)/p
=> k = -4 and p = 2h-4
Check the options given.
Hence option c is the answer.
Question 5: The circle C1: x2+y2= 3, with centre at O, intersects the parabola x2 = 2y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2√3 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the y-axis, then
(a) Q2 Q3 = 12
(b) R2R3 = 4√6
(c) Area of the triangle OR2 R3 is 6√2
(d) Area of the triangle PQ2 Q3 is 4√2
Solution:
Given circle, C1: x2+y2= 3 ..(i)
And parabola, x2 = 2y ..(ii)
y2+2y-3 = 0
(y+3)(y-1) = 0
=> y = 1 (since y not equal -3)
So x2 = 2
=> x = √2
=> P(√2, 1)
Equation of the tangent to the circle C1 at P is √2x + y – 3 = 0
Let the centre of circle C2 be (0, k).
=> r = 2√3
|(k-3)/√3| = 2√3
=>|k-3| = 6
=> k = 9 or -3
So Q2 (0, 9), Q3(0,-3)
Q2Q3 = √(0+12)2 = 12 (distance formula)
R2R3 = length of transverse common tangent
= √[(Q2Q3)2-(r1+r2)2]
= √(122 – 4√32)
= √96
= 4√6
Area of triangle OR2 R3 = ½ R2R3 × length of the perpendicular from origin to tangent
= ½ ×4√6×√3
= 6√2
Area of triangle PQ2 Q3 = ½ Q2Q3 × length of the perpendicular from y axis to P
= ½ ×12×√2
= 6√2
Hence option a, b, and c are correct.
Question 6: Let L be a normal to the parabola y2 = 4x. If L passes through the point (9, 6) then L is given by
(a) y-x+3 = 0
(b) y+3x-33 = 0
(c) y+x-15 = 0
(d) y-2x+12 = 0
Solution:
Equation of the normal is y = mx-2m-m3 ..(i)
Since the normal passes through (9, 6)
=> 6 = 9m – 2m – m3
=> m3 – 7m + 6 = 0
=> (m-1)(m-2)(m+3) = 0
=> m = 1, 2, -3
Put m = 1
=> y = x – 2 – 1
=> y-x+3 = 0
Put m = 2 in (i)
=> y = 2x – 4 – 8
=> y – 2x + 12 = 0
Put m = -3 in (i)
=> y = -3x + 6 + 27
=> y + 3x – 33 = 0
Hence option a, b, and d are correct.
Question 7: Let A and B be two distinct points on the parabola y2 = 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
Solution:
Let A(t12, 2t1) and B(t22, 2t2) be the points on the parabola y2 = 4x.
Its axis is the x axis.
The centre of the circle drawn with AB as the diameter is [(t12+t22)/2, (t1+t2)].
Since the circle meets axis of parabola,
r = |t1 + t2|
= ±r
Slope of AB = 2(t2-t1)/(t22-t12)
= 2/(t1+t2)
= ±2/r
Hence option c and d are correct.
Question 8: Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a>0. Length of the chord PQ is
(a) 7a
(b) 5a
(c) 2a
(d) 3a
Solution:
Let P(at2, 2at) and Q(a/t2, -2a/t) be the endpoints of a focal chord of the parabola y2 = 4ax.
Point of intersection of tangent at P and Q is (-a, a(t-1/t))
Point of intersection lie on the line y = 2x + a
=> a(t-1/t) = -2a + a
=> t-1/t = -1
(t+1/t)2 = (t-1/t)2 + 4
= 1 + 4
= 5
PQ = √((at2-a/t2)2 + (2at + 2a/t)2)
= a√((t+1/t)2 (t-1/t)2 + 4(t+1/t)2)
= a(t+1/t)√((t-1/t)2 + 4)
= a(t+1/t)2
= 5a
Hence option b is the answer.
Question 9: The equation of the directrix of the parabola y2+4y+4x+2 = 0 is
(a) x = -1
(b) x = 1
(c) x = -3/2
(d) x = 3/2
Solution:
Given equation of the parabola is y2+4y+4x+2 = 0
=> y2 + 4y + 4 = -4x + 2
=> (y+2)2 = -4(x-½)
This is of the form Y2 = -4AX
Equation of directrix is X = A
=> x-½ = 1
=> x = 1+ ½
=> x = 3/2
Hence option d is the answer.
Question 10: The angle between the tangents drawn from the point (1, 4) to the parabola y2 = 4x is
(a) π/6
(b) π/4
(c) π/3
(d) π/2
Solution:
Let m be the slope of the tangent to the parabola, then its equation is
y = mx + 1/m
Since the tangent passes through (1, 4)
4 = m + 1/m
=> m2 – 4m + 1 = 0
=> m1 + m2 = 4 (sum of roots)
m1m2 = 1 (product of roots)
If angle between two tangents be θ.
tan θ = |(m1 – m2)/(1+m1m2)|
= √((m1+m2)2 – 4m1m2)/(1+m1m2)
= √(16 – 4)/(1+1)
= √12/2
= √3
=> θ = π/3
Hence option c is the answer.
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