JEE Previous Year Questions With Solutions on Parabola

The parabola is the locus of a point which moves so that its distances from a fixed point, the focus, and a fixed straight line, the directric, are always equal. In conic section, parabola is one of the most important concepts. About 2-3 questions asked every year from this section in JEE exam. Practicing JEE previous year papers will help the JEE aspirants to understand the questions pattern, marking scheme and exam paper difficulty level. We have come up with the JEE Maths chapter-wise past year questions with solutions, practice papers and many more to help JEE aspirants during their preparation. Students are advised to practice all the questions below for scoring high in the conic section.

The PDF below consists of JEE past year questions with solutions on topic Parabola:

Download Parabola Solved Previous Year Questions PDF

JEE Maths Past Year Questions With Solutions on Parabola

Question 1 : If y = mx + 4 is a tangent to both the parabolas, y2 = 4a and x2 = 2by, then b is equal to

(a) -64                (b) 128                    (c) -128                   (d) -32

Answer: (c)

Solution:

Any tangent to parabola y2 = 4x is y = mx + a/m

On comparing with y = mx + 4, we have

1/m = 4 => m = 1/4

Now, equation of tangent y = x/4 + 4 to x2 = 2by

=> x2 = 2b(x/4 + 4)

or 2x2 – bx – 16b = 0

Find the value of b:

Discriminant, D = 0

so, b2 + 128 b = 0

=> b(b + 128) = 0

b = 0 or b = -128

=> b = 0 (not possible) so, b = -128. Answer!

Question 2: Find the shortest distance between the line x – y = 1 and the curve x = y2.

Solution:

x = y2

Differentiate above equation w.r.t. x to find the slope

1 = 2y dy/dx

=> dy/dx = 1/2y

1/2y = 1 => y = 1/2

and x = (1/2) 2 = 1/4

(1/4, 1/2) is point on a parabola.

Therefore, the shortest distance between line and curve is

1412+11+1\frac{|\frac{1}{4} – \frac{1}{2} + 1|}{\sqrt{1+1}}

= 3/4√2 or 3√2/8

Question 3: Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are

(a) x = ± (y + 2a)

(b) y = ± (x + 2a)

(c) x = ±(y + a)

(d) y = ±(x + a)

Answer: (d)

Solution: Let y = mx + 2a/m be equation of tangent to any parabola y2 = 8ax

If above tangent is tangent to the circle, x2 + y2 = 2a2, then

2a=±2amm2+1\sqrt{2}a= \pm \frac{2a}{m\sqrt{m^2+1}}

=>m2(1 + m2) = 2

=> (m2 + 2)(m2 – 1) = 0

=>m = ±1

So, required equation of tangent is y = ±(x + 2a)

Question 4: The tangent PT and the normal PN to the parabola y2 = 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) latus rectum is 2a/3

(d) focus is (a, 0)

Answer: (a) and (d)

Solution:

Let P(at2, 2at) be any point on the parabola y2 = 4ax.

Find the equation of tangent and normal:

Equation of tangent: y = x/t + at or yt = x + at2 and

Equation of normal: y = -tx + 2at + at2

Let centroid of triangle PTN is S(h, k)

Therefore,

h=at2+(at2)+2a+at3h = \frac{at^2 + (-at^2) + 2a + at}{3} and

k = 2at/3

=> 3h = 2a + 9k2/4a

=> 9k2 = 4a(3h -2a)

Latus rectum = 4a/3

Locus of centroid is y2 = 4a/3(x – 2a/3) and vertex (2a/3, 0)

Directrix: x – 2a/3 = -a/3 => x = a/3

So, the focus is (a/3 + 2a/3, 0) = > (a, 0)

Question 5: Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of PQ is

(a) y2 – 4x + 2 = 0

(b) y2 + 4x – 2 = 0

(c) x2 + 4y + 2 = 0

(d) x2 – 4y + 2 = 0

Answer: (a)

Solution:

Given point is P = (1, 0) and let another point Q = (h, k)

Such that, k2 = 8h

Let (x, y) be the midpoint of PQ

So, x = (h + 1)/2 and y = (k+0)/2

=> x = 2m – 1 and y = 2n

Now, (2y) 2 = 8(2x – 1)

=> y2 = 4x – 2

=> y2 – 4x + 2 = 0, which is locus of mid point PQ.

Question 6: The common tangents to the circle x2 + y2 = 2 and the parabola y2 = 8x touch the circle at the point P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is

(A) 3                     (B) 6                         (C) 9                        (D) 15

Answer: (d)

Solution:

Circle x2 + y2 = 2 and the parabola y2 = 8x (Given)

Parabola JEE Past years questions

Equation of tangent to parabola, y = mx + a/m

In this case a = 2

So, y = mx + 2/m

Now,

00+2/m1+m2=2\frac{|0-0+2/m|}{\sqrt{1+m^2}} = \sqrt{2}

=> 2 = m2 (1 + m2)

=> m = ± 1

Now equation of TP: -x + y = 2

So, P(-1, 1) and Q(-1, -1) and R (2,4) and S(2, -4)

Now, the area of the quadrilateral PQRS = 1/2 x (PQ + RS) × (Distance between PQ and RS)

= 1/2 x (2 + 8) × (1 + 2)

= 1/2 x 10 × 3

= 15 sq. units.

Question 7: The normal at the point (bt12, 2bt1) on a parabola meets the parabola again in the point (bt22, 2bt2) then

(a) t2 = -t1 – 2/t1

(b) t2 = -t1 + 2/t

(c) t2 = t1 – 2/t1

(d) t2 = t1 + 2/t1

Answer: (a)

Solution:

Equation of the normal to a parabola y2 = 4bx at point (bt12, 2bt1) is

y = -t1 x + 2bt1 + bt13

We per the question, it passes through the point (bt22, 2bt2), then

2bt2 = -t1bt22 + 2bt1 + bt13

2t2 – 2t1 = -t1 (t22 – t12) = -t1(t2 – t1) (t2 + t1)

=> 2 = -t1(t2 + t1)

=>(t2 + t1) = -2/t1

Or t2 = -t1 – 2/t1

Question 8: A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at

(a) (0, 2)                     (b) (1, 0)                        (c) (0, 1)                           (d) (2, 0)

Answer: (b)

Solution:

Vertex is mid point of origin (0, 0) and focus (2,0)

Vertex of the parabola = (0+22,0+02)(\frac{0+2}{2}, \frac{0+0}{2}) = (1, 0)

Question 9: If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2 = 4ax and x2 = 4ay then

(a)d2 + (2b + 3c)2 = 0

(b) d2 + (3b + 2c)2 = 0

(c) d2 + (2b – 3c)2 = 0

(d) d2 + (3b – 2c)2 = 0

Solution:

Here (0, 0) and (4a, 4a) are the points of intersection of given parabolas.

=> Equation of line passing through these points is y = x.

On comparing y = x with given line 2bx + 3cy + 4d = 0, we get

d = 0 and 2b + 3c = 0

=> (2b + 3c)2 + d2 = 0

Question 10: The locus of the vertices of the family of parabolas y=a3x33+a2x22ay = \frac{a^3x^3}{3} + \frac{a^2x}{2} – 2a is

(a) xy = 105/64

(b) xy = ¾

(c) xy = 35/16

(d) xy = 64/105

Answer: (a)

Solution:

The family of parabolas y=a3x33+a2x22ay = \frac{a^3x^3}{3} + \frac{a^2x}{2} – 2a

Let (x, y) be the vertex of the parabola.

x = a2/22a3/3=34a\frac{-a^2/2}{2a^3/3} = \frac{-3}{4a} and

y = (a44+4×a33×2a)4×4a33\frac{-(\frac{a^4}{4} + 4 \times \frac{a^3}{3} \times 2a)}{4 \times \frac{4a^3}{3}}

= (14+83)×a44a33\frac{-(\frac{1}{4} + \frac{8}{3}) \times a^4}{\frac{4a^3}{3}}

= -35/12 x a/4 x 3

= -35/16 a

Because, vertex of y = ax2 + bx + c is (b2a,(b24ac)4a)(\frac{-b}{2a}, \frac{-(b^2-4ac)}{4a} )

Now, xy = (-3/4a ) ( -35/16)a = 105/64

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