Hyperbola MCQ Quiz - Objective Question with Answer for Hyperbola - Download Free PDF

Concept:

If the equation of a hyperbola is, then the coordinates of foci will be (ae,0) and (-ae,0), where e is the eccentricity

and b² = a²(e² - 1)

Calculation:

Let the equation of hyperbola be 

then the coordinates of foci will be (ae,0) and (-ae,0), where e is the eccentricity

Distance between foci = 2ae = 16

⇒ 2a() = 16

⇒ a = 4

and b2 = a2(e2 - 1) = 32(2 - 1) = 32

⇒ b = 4

So the equation of the hyperbola is x2 − y2 = 32

∴ The correct option is (4).

Concept:

The equation of the tangent to the hyperbola  is

 where m is the slope of tangent.

Calculation:

Given hyperbola 4x2 - 5y2 = 20 

⇒  ⇒ a =  and b = 2

Since, the tangent is parallel to x - y = 2

⇒ The slope of tangent = m = 1

So The equation of tangent is

  y = x ± 

⇒ y = x ± 1

∴ The correct answer is option (3).

Calculation

4x² – 9y² = 36 ... (2)

Solve (1) & (2)

4x² – 9 (8x – x²) = 36

13x² –72x –36 = 0

(13x + 6) (x = 6) = 0

y → Imaginary 

 as P lies on 2x – 3y + y = 0

Centroid (h, k)

⇒ 

6h – 2y = 9k – 4

6x – 9y = 20 

Hence option 4 is correct

Calculation

Given

 and \(\mathrm{H}_{2}:-\frac{\mathrm{x}^{2}}{\mathrm{~A}^{2}}+\frac{y^{2}}{\mathrm{~B}^{2}}=1\)  

Latus rectum of H₁ = 

⇒  ...(1)

⇒  ...(2)

From (1) and (2)

a = 5√2 

b = 5√3

Latus rectum of H₂ =  

⇒ 

Product of the lengths of their transverse axes is , 

⇒ 

⇒ 

⇒ B = 5√5

⇒ A = 5√6

⇒ 

⇒ 

Calculation

be = 13, b = 5

a² = b² (e² – 1)

= b² e² – b²

= 169 – 25 = 144

Hence option 3 is correct

Concept:

Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)

• Length of Latus rectum = 

Calculation:

Given: 

Compare with the standard equation of a hyperbola: 

So, a2 = 100 and b2 = 75

∴ a = 10

Length of latus rectum =  = 

CONCEPT:

The properties of a rectangular hyperbola  are:

• Its centre is given by: (0, 0)
• Its foci are given by: (- ae, 0) and (ae, 0)
• Its vertices are given by: (- a, 0)  and (a, 0)
• Its eccentricity is given by: 
• Length of transverse axis = 2a and its equation is y = 0.
• Length of conjugate axis = 2b and its equation is x = 0.
• Length of its latus rectum is given by: 

CALCULATION:

Here, we have to find the equation of hyperbola whose length of latus rectum is 4 and the eccentricity is 3.

As we know that, length of latus rectum of a hyperbola is given by 

⇒ 

⇒ b² = 2a

As we know that, the eccentricity of a hyperbola is given by 

⇒ a2e2 = a2 + b2

⇒ 9a² = a² + 2a

⇒ a = 1/4

∵ b2 = 2a

⇒ b2 = 1/2

So, the equation of the required hyperbola is 16x² - 2y² = 1

Hence, option B is the correct answer.

Concept

The equation of the hyperbola is  

The distance between the foci of a hyperbola = 2ae 

Again, 

Calculations:

The equation of the hyperbola is  ....(1)

The distance between the foci of a hyperbola is 16 and its eccentricity e = √2.

We know that The distance between the foci of a hyperbola = 2ae 

⇒ 2ae = 16 

⇒ a  =   = 

Again, 

⇒ 

⇒ 

Equation (1) becomes

⇒ 

⇒ x2 - y2 = 32

Concept:

Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)

Calculation:

Given:

16x² – 9y² = 1

Compare with 

∴ a² = 1/16 and b² = 1/9

Eccentricity =  

Concept: 

Equation of hyperbola ,  

Eccentricity, e =   

Directrix, x  =  

Equation of hyperbola ,   

Eccentricity, e =  

Directrix, y =   

Calculation: 

Given hyperbolic equation are ,  3y2 - 2x2 = 12 

⇒  

On comparing with standard equation, a =  and  b = 2 .

We know that eccentricity, e =  

⇒ e =  

⇒ e =         

As we know that Directrix , y =  

∴ Directrix, y =  

⇒ Directrix, y =  . 

The correct option is 4 . 

Concept:

Standard equation of an hyperbola:  (a > b)

• Coordinates of foci = (± ae, 0)
• Eccentricity (e) =  ⇔ a2e2 = a2 + b2
• Length of Latus rectum = 

Calculation:

Given: 

Compare with the standard equation of a hyperbola: 

So, a2 = 16 and b2 = 9

⇒ a = 4 and b = 3   (a > b)

Now, Eccentricity (e) =  

= 

= 

= 

Coordinates of foci = (± ae, 0) 

= (± 5, 0)

Concept:

For the standard equation of a hyperbola,

Coordinates of foci = (± ae, 0)

Coordinates of vertices = (±a, 0)

Eccentricity, e = 

Equation of directrix, x = ± a/e

Calculation:

Given:

Compare with standard equation, we get

a² = 16 and b² = 9

Eccentricity e =

Now, Equation of directrix,

Concept:

The general equation of the hyperbola is:

Here, coordinates of foci are (±ae, 0).

And eccentricity = 

Calculation:

The equation 25x2 - 4y2 = 100 can be written as

This is the equation of a hyperbola.

On comparing it with the general equation of hyperbola, we get

⇒ a² = 4 and b² = 25

Now, the eccentricity is given by

Hence, the eccentricity is .

Concept:

The eccentricity 'e' of the hyperbola  is given by e = , for a > b.

Calculation:

Let's say that the equation of the hyperbola is .

Eccentricity is √2.

⇒ √2 = 

Squaring both sides, we get

⇒ 2 = 

⇒  = 1

⇒ a² = b²

The required equation is therefore,  or x2 - y2 = a2.

Concept:

For the standard equation of a hyperbola,

Coordinates of foci (± ae, 0)

Eccentricity 

Calculation:

Here, foci = (±3, 0) and eccentricity, 

∴ ae = 3 and 

So, the equation of the required hyperbola is