## Mathematics MCQ on Conic Section for JEE and Engineering Exam 2022

MCQ on Conic Section: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject. In this post we are providing you MCQ on Conic Section, which will be beneficial for you in upcoming JEE and Engineering Exams.

## MCQ on Conic Section

Q1. If 4x2 + py2 = 45 and x2 − 4y2 = 5 cut orthogonally, then the value of p is ______.
a) 8
b) 0
c) 9
d) 1

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

Q2. The centre of the circle passing through the point (0, 1) and touching the curve y = x2 at (2, 4) is ____________.
a) (−16 / 10, 53 / 5)
b) (−16 / 5, 53 / 10)
c) (16 / 5, -53 / 10)
d) none

Solution:

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)

Q3. The foci of the ellipse 25 (x + 1)2 + 9 (y + 2)2 = 225 are at ___________.
a) (1, 2); (1, 6)
b) (1, 2); (1, 4)
c) (1, 1); (1, 6)
d) none

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)

Q4. 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
d) none

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

Q5. The equation of 2x2 + 3y2 − 8x − 18y + 35 = k represents a ________.
a) -1
b) 0
c) 1
d) none

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.

Q6.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) 6
b) 7
c) 5
d) 8

Solution:

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

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

e\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 b= 16 (1 − [9 / 16]) = 7

Q7. On the parabola y = x2, the point least distance from the straight line y = 2x − 4 is ___________.
a) (-1,1)
b) (1,1)
c) (1,-1)
d) none

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).

Q8.  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.

Q9. The line x − 1 = 0 is the directrix of the parabola, y− kx + 8 = 0. Then, one of the values of k is ________.
a) −8, 4
b)  8,-4
c) -4,0
d) 0,8

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

Q10.  The eccentricity of the curve represented by the equation x+ 2y2 − 2x + 3y + 2 = 0 is __________.
a) e = 1 / 2
b) e = 1 / √2
c) e = √2
d) none

Solution:

Equation x+ 2y2 − 2x + 3y + 2 = 0 can be written as (x − 1)2 / 2 + (y + 3 / 4)= 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)

e= 1 − 1 / 2

e = 1 / √2

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