Maths Conic Sections Previous Year Questions With Solutions of IIT JEE & Engineering Exam

a) (1,1)

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).

d) 9

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

a) 7

d) P lies inside C but outside E

c) -8, 4

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

c) (1, 2); (1, 6)

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)

a) x + 1

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 + \sqrt{a^2m^2 − b^2}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 + \sqrt{3 \times (1)^2 − 2}3×(1)2 −2​

y = x + 1

d) (−16 / 5, 53 / 10)

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)

d) 0

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.

b) 0 

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.