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

Calculation:

Given,

Hyperbola equation: $25x^2-75y^2=225$

Divide both sides by 225 to obtain standard form:

$\frac{25x^2}{225}-\frac{75y^2}{225}=1⟹\frac{x^2}{9}-\frac{y^2}{3}=1$

Thus, $a^2=9$ and $b^2=3$.

Compute $c$ from $c^2=a^2+b^2$:

$c^2=9+3=12⟹c=\sqrt{12}=2\sqrt{3}.$

The foci are at $(\pm c,0)$, so the distance between them is $2c$:

$2c=2\times 2\sqrt{3}=4\sqrt{3}.$

∴ The distance between the two foci is $4\sqrt{3}$ 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.

Concept:

Hyperbola parameters:

• Given one focus of hyperbola $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is at $(\sqrt{10},0)$.
• The corresponding directrix is $x=\frac{9}{\sqrt{10}}$.
• Recall the relationships: $ae=c$ (distance of focus from origin), $\frac{a}{e}$ is directrix, and $c^2=a^2+b^2$.
• The length of the latus rectum is $l=\frac{2b^2}{a}$.

Calculation:

Given:

$ae=\sqrt{10}$ and $\frac{a}{e}=\frac{9}{\sqrt{10}}$

Then,

$a^2=9$ and $e=\frac{\sqrt{10}}{3}$

Now,

$(ae{)}^2=a^2+b^2⟹10=9+b^2⟹b^2=1$

$l=\frac{2b^2}{a}=\frac{2\times 1}{3}=\frac{2}{3}$

Therefore,

$9(e^2+l)=9({\left(\frac{\sqrt{10}}{3}\right)}^2+\frac{2}{3})=9(\frac{10}{9}+\frac{2}{3})=10+6=16$

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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

Distance between foci = 2ae = 16

⇒ 2a($\sqrt{2}$) = 16

⇒ a = 4$\sqrt{2}$

and b2 = a2(e2 - 1) = 32(2 - 1) = 32

⇒ b = 4$\sqrt{2}$

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

∴ The correct option is (4).

Concept:

The equation of the tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

$y=mx\pm \sqrt{a^2{m}^2-b^2}$ where m is the slope of tangent.

Calculation:

Given hyperbola 4x2 - 5y2 = 20 

⇒ $\frac{x^2}{5}-\frac{y^2}{4}=1$ ⇒ a = $\sqrt{5}$ 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 ± $\sqrt{5-4}$

⇒ y = x ± 1

∴ The correct answer is option (3).

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 = $\frac{2{\mathrm{b}}^2}{\mathrm{a}}$

Calculation:

Given: $\frac{{\mathrm{x}}^2}{100}-\frac{{\mathrm{y}}^2}{75}=1$

Compare with the standard equation of a hyperbola: $\frac{{\mathbf{x}}^2}{{\mathbf{a}}^2}-\frac{{\mathbf{y}}^2}{{\mathbf{b}}^2}=1$

So, a2 = 100 and b2 = 75

∴ a = 10

Length of latus rectum =  $\frac{2{\mathrm{b}}^2}{\mathrm{a}}$= $\frac{2\times 75}{10}=15$

CONCEPT:

The properties of a rectangular hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ 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: $e=\frac{\sqrt{a^2+b^2}}{a}$
• 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: $\frac{2b^2}{a}$

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 $\frac{2b^2}{a}$

⇒ $\frac{2b^2}{a}=4$

⇒ b² = 2a

As we know that, the eccentricity of a hyperbola is given by $e=\frac{\sqrt{a^2+b^2}}{a}$

⇒ 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 ${\displaystyle \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}}-{\displaystyle \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}}=1$ 

The distance between the foci of a hyperbola = 2ae 

Again, ${\mathrm{b}}^2={\mathrm{a}}^2({\mathrm{e}}^2-1)$

Calculations:

The equation of the hyperbola is ${\displaystyle \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}}-{\displaystyle \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}}=1$ ....(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  =  ${\displaystyle \frac{16}{2\sqrt{2}}}$ = $4\sqrt{2}$

Again, ${\mathrm{b}}^2={\mathrm{a}}^2({\mathrm{e}}^2-1)$

⇒ ${\mathrm{b}}^2=32(2-1)$

⇒ ${\mathrm{b}}^2=32$

Equation (1) becomes

⇒ ${\displaystyle \frac{{\mathrm{x}}^2}{32}}-{\displaystyle \frac{{\mathrm{y}}^2}{32}}=1$

⇒ 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

$\Rightarrow \frac{{\mathrm{x}}^2}{\frac{1}{16}}-\frac{{\mathrm{y}}^2}{\frac{1}{9}}=1$

Compare with $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

∴ a² = 1/16 and b² = 1/9

Eccentricity = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{1+\frac{\left(\frac{1}{9}\right)}{\left(\frac{1}{16}\right)}}=\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac{5}{3}$ 

Concept: 

Equation of hyperbola , $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ 

Eccentricity, e = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$  

Directrix, x  = $\pm \frac{\mathrm{a}}{\mathrm{e}}$ 

Equation of hyperbola , $-\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$  

Eccentricity, e = $\sqrt{1+\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}}$ 

Directrix, y = $\pm \frac{\mathrm{b}}{\mathrm{e}}$  

Calculation: 

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

⇒ $-\frac{{\mathrm{x}}^2}{6}+\frac{{\mathrm{y}}^2}{4}=1$ 

On comparing with standard equation, a = $\sqrt{6}$ and  b = 2 .

We know that eccentricity, e = $\sqrt{1+\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}}$ 

⇒ e = $\sqrt{1+\frac{6}{4}}$ 

⇒ e = $\frac{\sqrt{10}}{2}$        

As we know that Directrix , y = $\pm \frac{\mathrm{b}}{\mathrm{e}}$ 

∴ Directrix, y = $\pm \frac{2}{\frac{\sqrt{10}}{2}}$ 

⇒ Directrix, y = $\pm \frac{4}{\sqrt{10}}$ . 

The correct option is 4 . 

Concept:

Standard equation of an hyperbola: $\frac{{\mathbf{x}}^2}{{\mathbf{a}}^2}-\frac{{\mathbf{y}}^2}{{\mathbf{b}}^2}=1$ (a > b)

• Coordinates of foci = (± ae, 0)
• Eccentricity (e) = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$ ⇔ a2e2 = a2 + b2
• Length of Latus rectum = $\frac{2{\mathrm{b}}^2}{\mathrm{a}}$

Calculation:

Given: $\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1$

Compare with the standard equation of a hyperbola: $\frac{{\mathbf{x}}^2}{{\mathbf{a}}^2}-\frac{{\mathbf{y}}^2}{{\mathbf{b}}^2}=1$

So, a2 = 16 and b2 = 9

⇒ a = 4 and b = 3   (a > b)

Now, Eccentricity (e) = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$ 

= $\sqrt{1+\frac{9}{16}}$

= $\sqrt{\frac{16+9}{16}}$

= $\frac{5}{4}$

Coordinates of foci = (± ae, 0) 

= (± 5, 0)

Concept:

For the standard equation of a hyperbola, $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

Coordinates of foci = (± ae, 0)

Coordinates of vertices = (±a, 0)

Eccentricity, e = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$

Equation of directrix, x = ± a/e

Calculation:

Given:

$\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1$

Compare with standard equation, we get

a² = 16 and b² = 9

Eccentricity e = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{16+9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}$

Now, Equation of directrix,

$x=\pm \frac{a}{e}=\pm \frac{4}{\left(\frac{5}{4}\right)}=\pm \frac{16}{5}$

Concept:

The general equation of the hyperbola is:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

And eccentricity = $e=\sqrt{1+\frac{b^2}{a^2}}$

Calculation:

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

$\frac{x^2}{4}-\frac{y^2}{25}=1$

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

$e=\sqrt{1+\frac{25}{4}}=\frac{\sqrt{29}}{2}$

Hence, the eccentricity is $\frac{\sqrt{29}}{2}$.

Concept:

The eccentricity 'e' of the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ is given by e = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$, for a > b.

Calculation:

Let's say that the equation of the hyperbola is $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$.

Eccentricity is √2.

⇒ √2 = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$

Squaring both sides, we get

⇒ 2 = $1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$

⇒ $\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$ = 1

⇒ a² = b²

The required equation is therefore, $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1$ or x2 - y2 = a2.

Concept:

For the standard equation of a hyperbola,$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Coordinates of foci (± ae, 0)

Eccentricity $e=\sqrt{1+\frac{b^2}{a^2}}$

Calculation:

Here, foci = (±3, 0) and eccentricity, $e=\frac{3}{2}$

∴ ae = 3 and $e=\frac{3}{2}\Rightarrow a=2$

$\therefore b^2=a^2\left(e^2-1\right)\Rightarrow b^2=4\left(\frac{9}{4}-1\right)=4\times \frac{5}{4}=5$

So, the equation of the required hyperbola is

$\frac{x^2}{4}-\frac{y^2}{5}=1$