JEE Previous Year Question Bank On Conic Section

Conic section is a curve obtained as the intersection of the surface of a cone with a plane. This article covers basic definitions and standard equations for origin, shifted parabola, ellipse and hyperbola, equation of tangents and normals to parabola, ellipse and hyperbola in various forms, geometrical properties of all the three curves, parametric forms and equations with auxiliary circles, various shortcuts and properties of all three curves. The conic section questions from the previous years of JEE Main are present in this page along with the detailed solution for each question. These questions include all the important topics and formulae.

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JEE Main Maths Conic Section Previous Year Questions With Solutions

Question 1: The equation of 2x² + 3y² − 8x − 18y + 35 = k represents a ________.

Solution:

Given equation, 2x² + 3y² − 8x − 18y + 35 – k = 0

Compare with ax² + by² + 2hxy + 2gx + 2fy + c = 0,we get

a = 2, b = 3, h = 0, g = −4, f = −9, c = 35 − k

Δ = abc + 2fgh − af²− bg²− ch² = 6 (35 − k) + 0 − 162 − 48 − 0

Δ = 210 − 6k − 210 = −6k;

Δ = 0, if k = 0

So, that given equation is a point if k = 0.

Question 2: The locus of the midpoint of the line segment joining the focus to a moving point on the parabola y² = 4ax is another parabola with the directrix ___________.

Solution:

Let midpoint be (h, k) and moving point be (at², 2at).

Let focus is (a, 0).

So h = [at² + a] /2, k = [2at + 0] / 2

=> t = k/a

So 2h = a(k/a)² + a

= (k²/a) + a

=> k² = 2ah – a²

=> k² = 2a(h – a/2)

Replace (h, k) by (x, y)

y² = 2a(x – a/2)

Directrix is (x − a/2) = -a/2

=> x = 0

So x = 0 is the directrix.

Question 3: On the parabola y = x², the point least distance from the straight line y = 2x − 4 is ___________.

Solution:

Given, parabola y = x² …..(i)

Straight line y = 2x − 4 …..(ii)

From (i) and (ii),

x²− 2x + 4 = 0

Let f (x) = x²− 2x + 4,

f′(x) = 2x − 2.

For least distance, f′(x) =0

⇒ 2x − 2 = 0

x = 1

From y = x², y = 1

So, the point least distant from the line is (1, 1).

Question 4: The line x − 1 = 0 is the directrix of the parabola, y²− kx + 8 = 0. Then, one of the values of k is ________.

Solution:

The parabola is y² = 4 * [k / 4] (x − [8 / k]).

Putting y = Y, x − (8/k) = X, the equation is Y² = 4 * [k/4] * X

The directrix is X + (k/4) = 0,i.e. x − (8/k) + (k/4) = 0.

But x − 1 = 0 is the directrix.

So, [8 / k] − (k / 4) = 1

⇒ k = −8, 4

Question 5: The centre of the circle passing through the point (0, 1) and touching the curve y = x² at (2, 4) is ____________.

Solution:

Tangent to the parabola y = x²at (2, 4) is [1 / 2] (y + 4) = x * 2 or

4x − y − 4 = 0

It is also a tangent to the circle so that the centre lies on the normal through (2, 4) whose equation is x + 4y = λ, where 2 + 16 = λ.

Therefore, x + 4y = 18 is the normal on which lies (h, k).

h + 4k = 18 ….. (i)

Again, distance of centre (h, k) from (2, 4) and (0, 1) on the circle are equal.

Hence, (h − 2)²+ (k − 4)² = h² + (k − 1)²

So, 4h + 6k = 19 …..(ii)

Solving (i) and (ii), we get the centre = (−16 / 5, 53 / 10)

Question 6: Find the equation of the axis of the given hyperbola x²/3 − y²/2 = 1 which is equally inclined to the axes.

Solution:

x²/3 − y²/2 = 1

Equation of tangent is equally inclined to the axis i.e., tan θ = 1 = m.

Equation of tangent y = mx + √(a²m² – b²)

Given equation is [x²/3] − [y²/2] = 1 is an equation of hyperbola which is of form [x²/a²] − [y²/b²] = 1.

Now, on comparing a²= 3, b² = 2

y = 1 * x + √(3-2)

y = x + 1

Question 7: If 4x² + py² = 45 and x² − 4y² = 5 cut orthogonally, then the value of p is ______.

Solution:

Slope of 1^st curve (dy / dx)_I= −4x / py

Slope of 2^nd curve (dy / dx)_II= x / 4y

For orthogonal intersection (−4x / py) (x / 4y) = −1

x² = py²

On solving equations of given curves x = 3, y = 1

p (1) = (3)² = 9

p = 9

Question 8: If the foci of the ellipse (x² / 16 )+ (y² / b² )= 1 and the hyperbola (x² / 144) − (y² / 81) = 1 / 25 coincide, then the value of b² is _______.

Solution:

Hyperbola is (x² / 144) − (y² / 81) = 1 / 25

a = √(144/25); b = √(81/25)

e₁= √(1 + b²/a²)

= √(1 + 81/144)

= 15/12

= 5/4

Therefore, foci = (ae₁, 0) = ([12 / 5] * [5 / 4], 0) = (3, 0)

Therefore, focus of ellipse = (4e, 0) i.e. (3, 0)

=> e = 3/4

use formula e² = 1 – (b²/a²)

Hence b²= 16 (1 − [9 / 16]) = 7

Question 9: Let E be the ellipse x² / 9 + y² / 4 = 1 and C be the circle x² + y² = 9. Let P and Q be the points (1, 2) and (2, 1), respectively. Then

A) Q lies inside C but outside E

B) Q lies outside both C and E

C) P lies inside both C and E

D) P lies inside C but outside E

Solution:

The given ellipse is [x² / 9] + [y² / 4] = 1. The value of the expression [x² / 9] + [y² / 4] – 1 is positive for x = 1, y = 2 and negative for x = 2, y = 1. Therefore, P lies outside E and Q lies inside E. The value of the expression x² + y² – 9 is negative for both the points P and Q. Therefore, P and Q both lie inside C. Hence, P lies inside C but outside E.

Hence option d is the answer.

Question 10: The equation of the director circle of the hyperbola [x² / 16] − [y² / 4] = 1 is given by ______.

Solution:

Equation of the director circle of hyperbola is x² + y² = a² − b². Here a² = 16, b² = 4

Therefore, x² + y² = 12 is the required director circle.

Question 11: If m₁ and m₂ are the slopes of the tangents to the hyperbola x² / 25 − y² / 16 = 1 which pass through the point (6, 2), then find the relation between the sum and product of the slopes.

Solution:

The line through (6, 2) is y − 2 = m (x − 6)

y = mx + 2 − 6m

Now from condition of tangency, (2 − 6m)²= 25m² − 16

36m² + 4 − 24m − 25m² + 16 = 0

11m²− 24m + 20 = 0

Obviously its roots are m₁ and m₂, therefore m₁ + m₂ = 24 / 11 and m₁m₂= 20 / 11.

Question 12: The eccentricity of the curve represented by the equation x²+ 2y² − 2x + 3y + 2 = 0 is __________.

Solution:

Equation x²+ 2y² − 2x + 3y + 2 = 0 can be written as (x − 1)² / 2 + (y + 3 / 4)²= 1 / 16

[(x − 1)²] / (1/8) + [(y + 3 / 4)²] / (1/16) = 1, which is an ellipse with a² = 1 / 8 and b² = 1 / 16

e = √(1 – b²/a²)

e²= 1 − 1 / 2

e = 1 / √2

Question 13: The foci of the ellipse 25 (x + 1)² + 9 (y + 2)² = 225 are at ___________.

Solution:

[25(x + 1)²/225] + [9(y + 2)²/225] = 1

Here, a = √[225 / 25] = 15 / 5 = 3

b = √[225/9] = 15 / 3 = 5

Major axis of the ellipse lies along the y-axis.

b² = a²(1 – e²)

e = √[1 − 9 / 25] = 4 / 5

Focus = (−1, −2 ± [15 / 3] * [4 / 5])

= (−1, −2 ± 4)

= (-1, 2); (-1, -6)

Question 14: The locus of a variable point whose distance from (2, 0) is 2/3 times its distance from the line x = −9 / 2, is __________.

Solution:

Let point P be (x₁, y₁) and Q(2, 0).

Given line => x = -9/2

=> 2x + 9 = 0

Let PM be distance from P to line 2x + 9 = 0.

So, PQ = (2/3)PM

= (2/3)(x₁ + 9/2)

(x₁ + 2)² + y²₁ = 4/9 (x₁ + 9 / 2)²

9 [x₁²+ y²₁+ 4x₁ + 4] = 4(x₁²+ (81 / 4) + 9x₁)

5x₁²+ 9y²₁= 45

(x₁² / 9) + (y²₁/ 5 ) = 1,

Locus of (x₁, y₁) is (x²/9) + (y²/5) = 1, which is the equation of an ellipse.

Question 15: The equation of the ellipse whose latus rectum is 8 and whose eccentricity is 1 / √2, referred to the principal axes of coordinates, is __________.

Solution:

Equation of ellipse = (x²/a²) + (y²/b²) = 1

Latus rectum = 2b²/a = 8

=> b² = 4a ..(i)

Given e = 1/√2

We know e² = 1 – (b²/a²)

1/2 = 1 – 4/a

Solving we get a = 8.

So b² = 4a = 32

So a²= 64, b² = 32

Hence, the required equation of ellipse is (x²/64) + (y²/32) = 1.

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