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 4x^{2} + py^{2} = 45 and x^{2} − 4y^{2} = 5 cut orthogonally, then the value of p is ______.
a) 8
b) 0
c) 9
d) 1
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
Q2. The centre of the circle passing through the point (0, 1) and touching the curve y = x^{2} 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 = 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)
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 x^{2}/3 − y^{2}/2 = 1 which is equally inclined to the axes.
a) x-1
b) x+1
c) 1
d) none
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 + $amb $
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 + $×()2 $
y = x + 1
Q5. The equation of 2x^{2} + 3y^{2} − 8x − 18y + 35 = k represents a ________.
a) -1
b) 0
c) 1
d) none
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.
Q6.If the foci of the ellipse x^{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 _______.
a) 6
b) 7
c) 5
d) 8
Solution:
Hyperbola is x^{2} / 144 − y^{2} / 8 = 1 / 25
a = $25144 , b=2581 $ and
e_{1 }= $+14481 =144225 =1215 =45 $
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
Q7. On the parabola y = x^{2}, 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 = 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).
Q8. 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.
Q9. The line x − 1 = 0 is the directrix of the parabola, y^{2 }− 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 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
Q10. The eccentricity of the curve represented by the equation x^{2 }+ 2y^{2} − 2x + 3y + 2 = 0 is __________.
a) e = 1 / 2
b) e = 1 / √2
c) e = √2
d) none
Solution:
Equation x^{2 }+ 2y^{2} − 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 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
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