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

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## Maths Conic Sections Previous Year Questions With Solutions

Maths Conic Sections Previous Year Questions With Solutions : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you Maths Conic Sections Previous Year Questions With Solutions.

## Maths Conic Sections Previous Year Questions With Solutions

Q1.  On the parabola y = x2, the point least distance from the straight line y = 2x − 4 is ________.
a) (1,1)
b) (2,1)
c) (1,4)
d) (0,1)

a) (1,1)

Solution:

Given, parabola y = x2  …..(i)

Straight line y = 2x − 4      …..(ii)

From (i) and (ii),

x− 2x + 4 = 0

Let f (x) = x− 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).

Q2.  If 4x2 + py2 = 45 and x2 − 4y2 = 5 cut orthogonally, then the value of p is ______.
a) 5
b) 7
c) 8
d) 9

d) 9

Solution:

Slope of 1st curve (dy / dx)= −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

Q3.  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 _______.
a) 7
b) 5
c) 9
d) 11

a) 7

Q4. 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

d) P lies inside C but outside E

Q5. The line x − 1 = 0 is the directrix of the parabola, y− kx + 8 = 0. Then, one of the values of k is ________.
a) -4, 8
b) 8, 4
c) -8, 4
d) 4, 2

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

Q6. The foci of the ellipse 25 (x + 1)2 + 9 (y + 2)2 = 225 are at ___________.
a) (2, 5); (4, 9)
b) (1, 3); (1, 5)
c) (1, 2); (1, 6)
d) (1, 4); (1, 3)

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)

Q7. Find the equation of the axis of the given hyperbola x2/3 −  y2/2 = 1 which is equally inclined to the axes.
a) x + 1
b) x – 1
c) 1+ x2
d) 1 – x2

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}

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 a= 3, b2 = 2

y = 1 * x + \sqrt{3 \times (1)^2 − 2}

y = x + 1

Q8. The centre of the circle passing through the point (0, 1) and touching the curve y = x2 at (2, 4) is __________.
a) (−53 / 10, 5 / 14)
b) (−10 / 17, 13 / 10)
c) (−5 / 16, 53 / 10)
d) (−16 / 5, 53 / 10)

d) (−16 / 5, 53 / 10)

Tangent to the parabola y = xat (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)+ (k − 4)2 = h2 + (k − 1)2

So, 4h + 6k = 19     …..(ii)

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

Q9. 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 ___________.
a) 1
b) 4
c) 2
d) 0

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.

Q10. The equation of 2x2 + 3y2 − 8x − 18y + 35 = k represents a ________.
a) 1
b) 0
c) 7
d) 3

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 − af− bg− 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.

Maths Conic Sections Previous Year Questions With Solutions