JEE Main Conic Section Important Questions

JEE Main Conic Section Important Questions - A Conic Section is a curve developed by a plane after the intersection of the surface of the cone. A Conic Section can be divided into four basic sections - Parabola, Hyperbola, Ellipse, and Circle. As JEE Main Conic Section is one of the most important sections in the Mathematics Paper, students are often inquisitive about the tentative questions that can be asked in the following year. Therefore, we have mentioned the important questions from previous years' question papers.

The National Testing Agency or NTA has released the JEE Main 2024 Exam Dates on the official website. JEE Main 2024 Exam date session 1 is January 24 to February 1, 2024, and JEE Main session 2 exam date is April 1 to 15, 2024. In this article, candidates can find the JEE Main Conic Section Important Questions. 

LATEST - JEE Main 2024 Daily Practice Questions for 22 September 2023

JEE Main Conic Section Important Questions

1: The equation of 2p2 + 3t2 − 8p − 18t + 35 = k represents a ________.

Answer -  

Given equation, 2p2 + 3t2 − 8p − 18t + 35 – k = 0

Compare with ap2 + bt2 + 2hpy + 2gp + 2ty + 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.

2: The locus of the midpoint of the line segment joining the focus to a moving point on the parabola p2 = 4cx is another parabola with the directrix ___________.

Answer -

Let the midpoint be (u, k) and the moving point be (ct2, 2ct).

Let focus is (c, 0).

So h = [ct2 + c] /2, k = [2ct + 0] / 2

=> t = k/c

So 2h = c(k/c)2 + c

= (k2/c) + c

=> k2 = 2ch – c2

=> k2 = 2c(h – c/2)

Replace (u, k) by (x, p)

p2 = 2c(x – c/2)

Directrix is (x − c/2)  = -c/2

=> x = 0

So x = 0 is the directrix.

3: On the parabola a = b2, the point least distance from the straight line a = 2b − 4 is ___________.

Answer -

Given, parabola b = x2  …..(i)

Straight line a = 2b − 4      …..(ii)

From (i) and (ii),

b2 − 2b + 4 = 0

Let f (b) = b2 − 2b + 4,

f′(b) = 2b − 2.

For least distance, f′(b) =0

⇒ 2b − 2 = 0

b = 1

From a = b2, a = 1

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

4: The line a − 1 = 0 is the directrix of the parabola, b2 − ka + 8 = 0. Then, one of the values of k is ________.

Answer -

The parabola is b2 = 4 * [k / 4] (a − [8 / k]).

Putting b = B, a − (8/k) = A, the equation is B2 = 4 * [k/4] * A

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

But a − 1 = 0 is the directrix.

So, [8 / k] − (k / 4) = 1

⇒ k = −8, 4

5: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 _______.

Answer -

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

6: Find the equation of the axis of the given hyperbola a2/3 −  b2/2 = 1 which is equally inclined to the axes.

Answer -

a2/3 −  y2/2 = 1

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

Equation of tangent b= ma + √(c2m2 – d2)

Given equation is [a2/3] − [b2/2] = 1 is an equation of hyperbola which is of the form [a2/c2] − [b2/d2] = 1.

Now, on comparing c2 = 3, d2 = 2

b = 1 * a + √(3-2)

b = a + 1

7: If 4a2 + pb2 = 45 and a2 − 4b2 = 5 cut orthogonally, then the value of p is ______.

Answer -

Slope of 1st curve (dy / da)I = −4a / pb

Slope of 2nd curve (dy / da)II= a / 4b

For orthogonal intersection (−4a / pb) (a / 4b) = −1

a2 = pb2

On solving equations of given curves a = 3, b = 1

p (1) = (3)2 = 9

p = 9

8: The center of the circle passing through the point (0, 1) and touching the curve b = a2 at (2, 4) is ____________.

Solution:

Tangent to the parabola b = a2 at (2, 4) is [1 / 2] (b + 4) = a * 2 or

4a − b − 4 = 0

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

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

h + 4k = 18  ….. (i)

Again, the distance of center (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 center = (−16 / 5, 53 / 10)

9: Let E be the ellipse a2 / 9 + b2 / 4 = 1 and C be the circle a2 + b2 = 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

Answer -

The given ellipse is [a2 / 9] + [b2 / 4] = 1. The value of the expression [a2 / 9] + [b2 / 4] – 1 is positive for a = 1, b = 2 and negative for a = 2, b = 1. Therefore, P lies outside E and Q lies inside E. The value of the expression a2 + b2 – 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.

10: The equation of the director circle of the hyperbola [a2 / 16] − [b2 / 4] = 1 is given by ______.

Answer -

The equation of the director circle of the hyperbola is a2 + b2 = c2 − d2.  Here c2 = 16, d2 = 4

Therefore, a2 + b2 = 12 is the required director circle.

Question 11: If m1 and m2 are the slopes of the tangents to the hyperbola a2 / 25 − b2 / 16 = 1 which passes through the point (6, 2), then find the relation between the sum and product of the slopes.

Answer -

The line through (6, 2) is b − 2 = m (a − 6)

y = mx + 2 − 6m

Now from state 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.

12: The eccentricity of the curve represented by the equation a2 + 2b2 − 2a + 3b + 2 = 0 is __________.

Answer -

Equation a2 + 2b2 − 2a + 3b + 2 = 0 can be written as (a − 1)2 / 2 + (b + 3 / 4)2 = 1 / 16

[(a − 1)2] / (1/8) + [(b + 3 / 4)2] / (1/16) = 1,  which is an ellipse with c2 = 1 / 8 and d2 = 1 / 16

e = √(1 – d2/c2)

e2 = 1 − 1 / 2

e = 1 / √2

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

Answer -

[25(a + 1)2/225] + [9(b + 2)2/225] = 1

Here, c = √[225 / 25] = 15 / 5 = 3

d = √[225/9] = 15 / 3 = 5

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

d2 = c2(1 – e2)

e = √[1 − 9 / 25] = 4 / 5

Focus = (−1, −2 ± [15 / 3] * [4 / 5])

= (−1, −2 ± 4)

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

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

Solution: 

Let point P be (a1, b1) and Q(2, 0).

Given line => a = -9/2

=> 2a + 9 = 0

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

So, PQ = (2/3)PM

= (2/3)(a1 + 9/2)

(a1 + 2)2 + b21 = 4/9 (a1 + 9 / 2)2

9 [a12 + b21+ 4a1 + 4] = 4(a12 + (81 / 4) + 9a1)

5a12 + 9b21 = 45

(a12 / 9) + (b21 / 5 ) = 1,

Locus of (a1, y1) is (a2/9) + (b2/5) = 1, which is the equation of an ellipse.

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 = (a2/c2) + (b2/d2) = 1

Latus rectum = 2d2/c = 8

=> d2 = 4c ..(i)

Given e = 1/√2

We know e2 = 1 – (d2/c2)

1/2 = 1 – 4/c

Solving we get c = 8.

So d2 = 4c = 32

So d2 = 64, c2 = 32

Hence, the required equation of the ellipse is (a2/64) + (b2/32) = 1.

JEE Main Mathematics Question Paper 

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