JEE Main Maths Conic Sections Previous Year Questions With Solutions

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:

α = [at2 + a] / 2, β = [2at + 0] / 2

⇒ 2α = at2 + a, at = β

2α = a * [β2 / a2] + a or

2aα = β2 / a2

The locus is y2 = 4a / 2 (x − [a / 2])

= 4b (x − b), (b = a / 2)

Directrix is (x − b) + b = 0 or x = 0.

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 − 8k = 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 = x2 at (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 + a2m2b2\sqrt{a^2m^2 − b^2}

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×(1)2 −2\sqrt{3 \times (1)^2 − 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 1st curve (dy / dx)I = −4x / py

Slope of 2nd 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 / 8 = 1 / 25 coincide, then the value of b2 is _______.

Solution: 

Hyperbola is x2 / 144 − y2 / 8 = 1 / 25

a = 14425,   b=8125\sqrt{\frac{144 }{ 25}}, \;  b = \sqrt{\frac{81 }{ 25}} and

e1 = 1+81144=225144=1512=54\sqrt{1 + \frac{81 }{ 144}} = \sqrt{\frac{225 }{ 144}} = \frac{15 }{ 12} = \frac{5 }{ 4}

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

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

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.

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

Therefore, 1 / 16 = 1 / 8 (1 − e2)

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,  b = √[225/9] = 15 / 3

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 (x1, y1)

So, (x1 +2)2 +y12\sqrt{(x_1 + 2)^2 + y^2_1} = 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:

Here, 2b2 a\frac{2b^2}{ a} = 8, e = 12\frac{1 }{\sqrt{2}}

a2 = 64, b2 = 32

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

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