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

Calculation:

Given,

Hyperbola equation:

Divide both sides by 225 to obtain standard form:

Thus, and .

Compute from :

The foci are at , so the distance between them is :

∴ The distance between the two foci is  units.

Hence, the correct answer is Option 2.

Calculation: 

Concept:

• Hyperbola Standard Equation: If the vertices are on the x-axis at (±a, 0), the equation is (x²/a²) - (y²/b²) = 1.
• Vertices: Given as (±6, 0), so a = 6 ⇒ a² = 36.
• Eccentricity: e = √5 / 2. For a hyperbola, e = √(1 + b²/a²).
• Normal: A line perpendicular to the tangent at a point on the curve. The slope of the normal is the negative reciprocal of the tangent slope.
• The given normal is parallel to the line √2x + y = 2√2, hence the slope is -√2.

Calculation:

Given:

a = 6 ⇒ a² = 36,

and e = √5 / 2

⇒ e² = 5 / 4

⇒ 5 / 4 = 1 + b² / 36 ⇒ b² = 9

So the hyperbola is: (x² / 36) - (y² / 9) = 1

Differentiating implicitly:

(2x / 36) - (2y / 9) × (dy/dx) = 0 ⇒ x / 18 = (2y / 9)(dy/dx)

⇒ dy/dx = x / 4y

⇒ slope of tangent = x / 4y

⇒ slope of normal = -4y / x

Given: slope of normal = -√2 ⇒ -4y / x = -√2 ⇒ 4y / x = √2

⇒ y = (x√2) / 4

Substitute into hyperbola:

(x² / 36) - (1 / 9) × ((x√2 / 4)²) = 1

⇒ (x² / 36) - (1 / 9) × (2x² / 16) = 1

⇒ (x² / 36) - (x² / 72) = 1

⇒ (x²)(1/36 - 1/72) = 1 ⇒ x² × (1/72) = 1 ⇒ x² = 72

⇒ x = 6√2

Then, y = (x√2) / 4 = (6√2 × √2) / 4 = 12 / 4 = 3

So, the point on the hyperbola is (6√2, 3)

The equation of the normal line with slope -√2 passing through (6√2, 3):

y - 3 = -√2(x - 6√2)

To find the y-intercept, put x = 0:

y = 3 + √2 × 6√2 = 3 + 12 = 15

So, line segment lies between (6√2, 3) and (0, 15)

Length d = √[(6√2)² + (15 - 3)²] = √[72 + 144] = √216

∴ The required value of d² is 216.

Calculation:

The hyperbola is 

With foci at $(\pm c,0)$, Where c² = a²+ b²

For P ( 4, 2√3)  the distances to the foci are 

⇒ 

Also D₁D₂ = 32 

⇒ ( 4 - )² + (2√3)². ( 4 + c)2 + (2√3)2 = 32² = 1024

⇒ (c²− 8c + 28) (c²+8c+28) = 1024

⇒ (c² + 28)² − (8c)² = 1024 

⇒ c⁴ − 8c² + 784 = 1024

⇒ c⁴ − 8c² − 240 = 0.

Let  u = c²

⇒ u² - 8u -240 = 0

⟹  

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⇒ c² = u = 20 ⟹ a² + b² = 20.

Since P Lies on hyperbola

⇒ 

use b²= 20 − a². Substitute

⇒ 16 (20 − a²) − 12a² = a²⁽20−a²⁾

⇒ 320 − 16a² − 12a² = 20a² − a^4,

⇒ a⁴ − 48a² + 320 = 0.

Let v =  a²

⇒ v² − 48v + 320 = 0 

If a² = 40, then  b^2=20-40=-20" id="MathJax-Element-186-Frame" role="presentation" style="position: relative;" tabindex="0"> b^2=20-40=-20" id="MathJax-Element-110-Frame" role="presentation" style="position: relative;" tabindex="0"> b^2=20-40=-20" id="MathJax-Element-1258-Frame" role="presentation" style="position: relative;" tabindex="0"> b^2=20-40=-20 $b^2=20-40=-20$ $b^2=20-40=-20$ $b^2=20-40=-20$ b² = 20 − 40 = −20 (impossible). Hence a² = 8 , b² = 12

Conjugate axis length p = 2b = 2√12 = 4√3 

⇒ p² = 48

and q² = 72 

Then, p² + q² = 48 +  72 = 120

Hence, the correct answer is 120.

Concept:

Hyperbola parameters:

• Given one focus of hyperbola is at .
• The corresponding directrix is .
• Recall the relationships: (distance of focus from origin), is directrix, and .
• The length of the latus rectum is .

Calculation:

Given:

and

Then,

and

Now,

Therefore,

Hence, the correct answer is Option 3.

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:

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