JEE > All Types of Questions for JEE > Single Correct MCQs: Conic Sections

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**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 y ^{2} = 4ax is(a) x = at; y = 2at(b) x = at^{2}; y = 2at(c) x = at^{2}; y^{2} = at^{3}(d) x = at^{2}; y = 4at**

Correct Answer is option (b)

The parametric equation of the parabola y^{2}= 4ax is x = at^{2}; y = 2at.

**Q.3. The centre of the circle 4x ^{2} + 4y^{2} – 8x + 12y – 25 = 0 is**

Correct Answer is option (b)

Given circle equation: 4x^{2}+ 4y^{2}– 8x + 12y – 25 = 0

⇒ x^{2}+ y^{2}– (8x/4) + (12y/4) – (25/4) = 0

⇒ x^{2}+y^{2}-2x +3y -(25/4) = 0 …(1)

As we know that the general equation of a circle is x^{2}+y^{2}+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 y ^{2}+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: y^{2}+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, Y^{2}= -4X …(1)

Hence, equation (1) is of the form y^{2}= -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 y^{2}+4y+4x+2=0 is 3/2.

**Q.5. The number of tangents that can be drawn from (1, 2) to x ^{2}+y^{2} = 5 is**

Correct Answer is option (b)

Given circle equation: x^{2}+y^{2}= 5

x^{2}+y^{2}-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 x ^{2} = -9y is equal to**

Correct Answer is option (d)

Given parabola equation: x^{2}= -9y …(1)

Since the coefficient of y is negative, the parabola opens downwards.

The general equation of parabola is x^{2}= -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 x^{2}= -9y is equal to 9 units.

**Q.7. For the ellipse 3x ^{2}+4y^{2} = 12, the length of the latus rectum is:**

Correct Answer is option (c)Given ellipse equation: 3x

^{2}+4y^{2}= 12

The given equation can be written as (x^{2}/4) + (y^{2}/3) = 1…(1)

Now, compare the given equation with the standard ellipse equation: (x^{2}/a^{2}) + (y^{2}/b^{2}) = 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 3x^{2}+4y^{2}= 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 y^{2}= 4ax …(2)

Comparing (1) and (2), we get

4a = 8

a = 8/4 = 2

We know that the focus of parabolic equation y^{2}= 4ax is (a, 0).

Therefore, the focus of the parabola y^{2}=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 b^{2}= a2(1-e^{2})

b^{2}= a^{2}– a^{2}e^{2}

b^{2}= a^{2}– (ae)^{2}

Now, substitute the values to find the value of a.

(4)^{2}= a^{2}-(3)^{2}

16 = a^{2}– 9

a^{2}= 16+9 = 25.

So, a = 5.The formula to calculate the eccentricity of ellipse is e = √[1-(b

^{2}/a^{2})]

e = √[1-(4^{2}/5^{2})]

e = √[(25-16)/25]

e = √(9/25) = 3/5.

Hence, option (c) 3/5 is the correct answer.

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