JEE Advanced (Single Correct Type): Conic Sections

# JEE Advanced (Single Correct Type): Conic Sections | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The length of the transverse axis is the distance between the ____.
(a) Two vertices
(b) Two Foci
(c) Vertex and the origin
(d) Focus and the vertex

Correct Answer is option (a)
The length of the transverse axis is the distance between two vertices.

Q.2. The parametric equation of the parabola y2 = 4ax is
(a) x = at; y = 2at
(b) x = at2; y = 2at
(c) x = at2; y2 = at3
(d) x = at2; y = 4at

Correct Answer is option (b)
The parametric equation of the parabola y2 = 4ax is x = at2; y = 2at.

Q.3. The centre of the circle 4x2 + 4y2 – 8x + 12y – 25 = 0 is
(a) (-2, 3)
(b) (1, -3/2)
(c) (-4, 6)
(d) (4, -6)

Correct Answer is option (b)
Given circle equation: 4x2 + 4y2 – 8x + 12y – 25 = 0
⇒ x2 + y2 – (8x/4) + (12y/4) – (25/4) = 0
⇒ x2 +y2 -2x +3y -(25/4) = 0 …(1)
As we know that the general equation of a circle is x2+y2+2gx+2fy+c=0, and the centre of the circle = (-g, -f)
Hence, by comparing equation (1) and the general equation,
2g = -2,
Thus, g = -1
2f = 3,  thus, f = 3/2
Now, substitute the values in the centre of the circle (-g, -f), we get,
Centre = (1, -3/2).
Therefore, option (b) (1, -3/2) is the correct answer.

Q.4. The equation of the directrix of the parabola y2+4y+4x+2=0 is
(a) x = 1
(b) x = -1
(c) x = 3/2
(d) x = -3/2

Correct Answer is option (c)
Given equation: y2+4y+4x+2=0
Rearranging the equation, we get
(y+2)2 = -4x+2
(y+2)2 = -4(x – (½))
Let Y = y+2 and X = x-(½)
So, Y2 = -4X …(1)
Hence, equation (1) is of the form y2 = -4ax. …(2)
By comparing (1) and (2), we get a=1.
We know that equation of directrix is x= a
Now, substitute a = 1 and x = x-(½) in the directrix equation
x – (½) = 1
x = 1+(½) = 3/2.
Therefore, the equation of the directrix of the parabola y2+4y+4x+2=0 is 3/2.

Q.5. The number of tangents that can be drawn from (1, 2) to x2+y2 = 5 is
(a) 0
(b) 1
(c) 2
(d) More than 2

Correct Answer is option (b)
Given circle equation: x2+y2= 5
x2+y2-5 =0 …(1)
Now, substitute (1, 2) in equation (1), we get
Circle Equation: (1)2+(2)2 -5 =0
Equation of circle = 1+5-5 =0
This represents that the point lies on the circumference of a circle, and hence only one tangent can be drawn from (1, 2).
So, option (b) 1 is the correct answer.

Q.6. The length of the latus rectum of x2 = -9y is equal to
(a) 3 units
(b) -3 units
(c) 9/4 units
(d) 9 units

Correct Answer is option (d)
Given parabola equation: x2 = -9y …(1)
Since the coefficient of y is negative, the parabola opens downwards.
The general equation of parabola is x2= -4ay…(2)
Comparing (1) and (2), we get

-4a = -9

a = 9/4
We know that the length of latus rectum = 4a = 4(9/4) = 9.
Therefore, the length of the latus rectum of x2 = -9y is equal to 9 units.

Q.7. For the ellipse 3x2+4y2 = 12, the length of the latus rectum is:
(a) 2/5
(b) 3/5
(c) 3
(d) 4

Correct Answer is option (c)

Given ellipse equation: 3x2+4y2 = 12
The given equation can be written as (x2/4) + (y2/3) = 1…(1)
Now, compare the given equation with the standard ellipse equation: (x2/a2) + (y2/b2) = 1, we get
a = 2 and b = √3
Therefore, a > b.
If a>b, then the length of latus rectum is 2b2/a
Substituting the values in the formula, we get
Length of latus rectum = [2(√3)2] /2 = 3
Therefore, the length of the latus rectum of the ellipse 3x2+4y2 = 12 is 3.

Q.8. The eccentricity of hyperbola is
(a) e =1
(b) e > 1
(c) e < 1
(d) 0 < e < 1

Correct Answer is option (b)

The eccentricity of hyperbola is greater than 1. (i.e.) e > 1.

Q.9. The focus of the parabola y2 = 8x is
(a) (0, 2)
(b) (2, 0)
(c) (0, -2)
(d) (-2, 0)

Correct Answer is option (b)

Given parabola equation y2 = 8x …(1)
Here, the coefficient of x is positive and the standard form of parabola is y2 = 4ax …(2)
Comparing (1) and (2), we get
4a = 8
a = 8/4 = 2
We know that the focus of parabolic equation y2 = 4ax is (a, 0).
Therefore, the focus of the parabola y2 =8x is (2, 0).
Hence, option (b) (2, 0) is the correct answer.

Q.10. In an ellipse, the distance between its foci is 6 and the minor axis is 8, then its eccentricity is
(a) 1/2
(b) 1/5
(c) 3/5
(d) 4/5

Correct Answer is option (c)

Given that the minor axis of ellipse is 8.(i.e) 2b = 8. So, b=4.

Also, the distance between its foci is 6. (i.e) 2ae = 6
Therefore, ae = 6/2 = 3
We know that b2 = a2(1-e2)
b2 = a2 – a2e2
b2 = a2 – (ae)2
Now, substitute the values to find the value of a.
(4)2 = a2 -(3)2
16 = a2 – 9
a2 = 16+9 = 25.
So, a = 5.

The formula to calculate the eccentricity of ellipse is e = √[1-(b2/a2)]
e = √[1-(42/52)]
e = √[(25-16)/25]
e = √(9/25) = 3/5.
Hence, option (c) 3/5 is the correct answer.

The document JEE Advanced (Single Correct Type): Conic Sections | Chapter-wise Tests for JEE Main & Advanced is a part of the JEE Course Chapter-wise Tests for JEE Main & Advanced.
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## Chapter-wise Tests for JEE Main & Advanced

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## Chapter-wise Tests for JEE Main & Advanced

447 docs|930 tests

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