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 2x^{2} + 3y^{2} − 8x − 18y + 35 = k represents a ________.

**Solution: **

Given equation, 2x^{2} + 3y^{2} − 8x − 18y + 35 – k = 0

Compare with ax^{2} + by^{2} + 2hxy + 2gx + 2fy + c = 0,we get

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

Δ = abc + 2fgh − af^{2 }− bg^{2 }− ch^{2} = 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 y^{2} = 4ax is another parabola with the directrix ___________.

**Solution:**

α = [at^{2} + a] / 2, β = [2at + 0] / 2

⇒ 2α = at^{2} + a, at = β

2α = a * [β^{2} / a^{2}] + a or

2aα = β^{2} / a^{2}

The locus is y^{2} = 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 = x^{2}, the point least distance from the straight line y = 2x − 4 is ___________.

**Solution:**

Given, parabola y = x^{2} …..(i)

Straight line y = 2x − 4 …..(ii)

From (i) and (ii),

x^{2 }− 2x + 4 = 0

Let f (x) = x^{2 }− 2x + 4,

f′(x) = 2x − 2.

For least distance, f′(x) =0

⇒ 2x − 2 = 0

x = 1

From y = x^{2}, 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, y^{2 }− kx + 8 = 0. Then, one of the values of k is ________.

**Solution:**

The parabola is y^{2} = 4 * [k / 4] (x − [8 / k]).

Putting y = Y, x − 8k = X, the equation is Y^{2} = 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 = x^{2} at (2, 4) is ____________.

**Solution:**

Tangent to the parabola y = x^{2 }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} = h^{2} + (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 x^{2}/3 − y^{2}/2 = 1 which is equally inclined to the axes.

**Solution:**

x^{2}/3 − y^{2}/2 = 1

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

Equation of tangent y = mx +

Given equation is [x^{2}/3] − [y^{2}/2] = 1 is an equation of hyperbola which is of form [x^{2}/a^{2}] − [y^{2}/b^{2}] = 1.

Now, on comparing a^{2 }= 3, b^{2} = 2

y = 1 * x +

y = x + 1

**Question 7:** If 4x^{2} + py^{2} = 45 and x^{2} − 4y^{2} = 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

x^{2} = py^{2}

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 ^{2} / 16 + y^{2} / b^{2} = 1 and the hyperbola x^{2} / 144 − y^{2} / 8 = 1 / 25 coincide, then the value of b^{2} is _______.

**Solution: **

Hyperbola is x^{2} / 144 − y^{2} / 8 = 1 / 25

a =

e_{1 }=

Therefore, foci = (ae_{1}, 0) = ([12 / 5] * [5 / 4], 0) = (3, 0)

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

Hence b^{2 }= 16 (1 − [9 / 16]) = 7

**Question 9:** Let E be the ellipse x^{2} / 9 + y^{2} / 4 = 1 and C be the circle x^{2} + y^{2} = 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 [x^{2} / 9] + [y^{2} / 4] = 1. The value of the expression [x^{2} / 9] + [y^{2} / 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 x^{2} + y^{2} – 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 [x^{2} / 16] − [y^{2} / 4] = 1 is given by ______.

**Solution: **

Equation of the director circle of hyperbola is x^{2} + y^{2} = a^{2} − b^{2}. Here a^{2} = 16, b^{2} = 4

Therefore, x^{2} + y^{2} = 12 is the required director circle.

**Question 11:** If m_{1} and m_{2} are the slopes of the tangents to the hyperbola x^{2} / 25 − y^{2} / 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 }= 25m^{2} − 16

36m^{2} + 4 − 24m − 25m^{2} + 16 = 0

11m^{2 }− 24m + 20 = 0

Obviously its roots are m_{1} and m_{2}, therefore m_{1} + m_{2} = 24 / 11 and m_{1}m_{2 }= 20 / 11.

**Question 12: **The eccentricity of the curve represented by the equation x^{2 }+ 2y^{2} − 2x + 3y + 2 = 0 is __________.

**Solution:**

Equation x^{2 }+ 2y^{2} − 2x + 3y + 2 = 0 can be written as (x − 1)^{2} / 2 + (y + 3 / 4)^{2 }= 1 / 16

^{2}] / (1/8) + [(y + 3 / 4)

^{2}] / (1/16) = 1, which is an ellipse with a

^{2}= 1 / 8 and b

^{2}= 1 / 16

Therefore, 1 / 16 = 1 / 8 (1 − e^{2})

e^{2 }= 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 (x_{1}, y_{1)}

So, _{1} + 9/2)

(x_{1} + 2)^{2} + y^{2}_{1} = 4/9 (x_{1} + 9 / 2)^{2}

9 [x_{1}^{2 }+ y^{2}_{1}+ 4x_{1} + 4] = 4(x_{1}^{2 }+ 81 / 4 + 9x_{1})

5x_{1}^{2 }+ 9y^{2}_{1 }= 45

x_{1}^{2} / 9 + y^{2}_{1 }/ 5 = 1,

Locus of (x_{1}, y_{1}) is x^{2}/9 + y^{2}/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,

a^{2 }= 64, b^{2} = 32

Hence, the required equation of ellipse is x^{2}/64 + y^{2}/32 = 1.

**Solve More Related Questions:**

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