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 \({x^2 \over a^2}-{y^2 \over b^2} =1\)

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 \({x^2\over a^2 }-{y^2 \over b^2} = 1\) is

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

Calculation:

Given hyperbola 4x2 - 5y2 = 20 

⇒ \({x^2\over 5 }-{y^2 \over 4} = 1\) ⇒ 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).

Calculation

\(x^{2}+y^{2}-8 x=0, ...(1)\)

\(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\)

4x² – 9y² = 36 ... (2)

Solve (1) & (2)

4x² – 9 (8x – x²) = 36

13x² –72x –36 = 0

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

\(x=\frac{-6}{13}, x=6\)

\(x=\frac{-6}{13}(\text { rejected })\)

y → Imaginary 

\(\frac{36}{9}-\frac{\mathrm{y}^{2}}{4}=1\)

\(\mathrm{y}^{2}=12, \mathrm{y}=\pm \sqrt{12}\)

\(\mathrm{A}(6, \sqrt{12}), \mathrm{B}(6,-\sqrt{12})\)

\(\mathrm{p}\left(\alpha, \frac{2 \alpha+4}{3}\right)\) as P lies on 2x – 3y + y = 0

Centroid (h, k)

\(\mathrm{h}=\frac{12+\alpha}{3}⇒ \alpha=3 \mathrm{~h}-12\)

\(\mathrm{k}=\frac{\frac{2 \alpha+4}{3}}{3} \Rightarrow 2 \alpha+4=9 \mathrm{k}\)

⇒ \(\alpha=\frac{9 \mathrm{k}-4}{2}\)

6h – 2y = 9k – 4

6x – 9y = 20 

Hence option 4 is correct

Calculation

Given

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

Latus rectum of H₁ = \(15 \sqrt{2}\)

⇒ \(\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=15 √{2}\) ...(1)

\(e_{1}=\sqrt{\frac{5}{2}}\)

⇒ \(1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{5}{2}\) ...(2)

From (1) and (2)

a = 5√2 

b = 5√3

Latus rectum of H₂ = \(12 \sqrt{5}\) 

⇒ \(\frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=12 √{5}\)

Product of the lengths of their transverse axes is \(100 \sqrt{10}\), 

⇒ \(2 \mathrm{a} \cdot 2 \mathrm{~B}=100 √{10}\)

⇒ \(2.5 √{2} .2 \mathrm{~B}=100 √{10}\)

⇒ B = 5√5

⇒ A = 5√6

\(\mathrm{e}_{2}^{2}=1+\frac{\mathrm{A}^{2}}{\mathrm{~B}^{2}}\)

⇒ \(1+\frac{150}{125}\)

⇒ \(\mathrm{e}_{2}^{2}=1+\frac{30}{25}\)

\(25 \mathrm{e}_{2}^{2}=55\)

Calculation

be = 13, b = 5

a² = b² (e² – 1)

= b² e² – b²

= 169 – 25 = 144

\(\ell(\mathrm{LR})=\frac{2 \mathrm{a}^{2}}{\mathrm{~b}}=\frac{2 \times 144}{5}=\frac{288}{5}\)

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 = \(\rm \frac{2b^2}{a}\)

Calculation:

Given: \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\)

Compare with the standard equation of a hyperbola: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)

So, a2 = 100 and b2 = 75

∴ a = 10

Length of latus rectum =  \(\rm \frac{2b^2}{a}\)= \(\rm \frac{2 \times 75}{10} = 15\)

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 \(\rm \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) 

The distance between the foci of a hyperbola = 2ae 

Again, \(\rm b^2 = a^2(e^2-1)\)

Calculations:

The equation of the hyperbola is \(\rm \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) ....(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  =  \(\dfrac {16}{2\sqrt 2}\) = \({4\sqrt 2}\)

Again, \(\rm b^2 = a^2(e^2-1)\)

⇒ \(\rm b^2 = 32(2-1)\)

⇒ \(\rm b^2 = 32\)

Equation (1) becomes

⇒ \(\rm \dfrac{x^2}{32} - \dfrac{y^2}{32} = 1 \)

⇒ 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

\( \Rightarrow \frac{{{\rm{\;}}{{\rm{x}}^2}}}{{\frac{1}{{16}}}} - \frac{{{{\rm{y}}^2}}}{{\frac{1}{9}}} = 1\)

Compare with \(\frac{{{\rm{\;}}{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\)

∴ a² = 1/16 and b² = 1/9

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

Concept: 

Equation of hyperbola , \(\rm \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1\) 

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

Directrix, x  = \(\rm\pm \frac{a}{e}\) 

Equation of hyperbola , \(\rm- \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1\)  

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

Directrix, y = \(\rm\pm \frac{b}{e}\)  

Calculation: 

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

⇒ \(\rm- \frac{x^{2}}{6}+\frac{y^{2}}{4}= 1\) 

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

We know that eccentricity, e = \(\rm\sqrt{1+\frac{a^{2}}{b^{2}}}\) 

⇒ e = \(\rm\sqrt{1+\frac{6}{4}}\) 

⇒ e = \(\frac{\sqrt{10}}{2}\)        

As we know that Directrix , y = \(\rm\pm \frac{b}{e}\) 

∴ 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: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) (a > b)

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

Calculation:

Given: \(\rm \frac{x^2}{16} - \frac{y^2}{9} = 1\)

Compare with the standard equation of a hyperbola: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)

So, a2 = 16 and b2 = 9

⇒ a = 4 and b = 3   (a > b)

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

= \(\sqrt {1 + \frac{9}{16}}\)

= \(\sqrt { \frac{16+9}{16}}\)

= \(\frac 54\)

Coordinates of foci = (± ae, 0) 

= (± 5, 0)

Concept:

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

Coordinates of foci = (± ae, 0)

Coordinates of vertices = (±a, 0)

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

Equation of directrix, x = ± a/e

Calculation:

Given:

\(\frac{{{{\rm{x}}^2}}}{{16}} - \frac{{{{\rm{y}}^2}}}{9} = 1\)

Compare with standard equation, we get

a² = 16 and b² = 9

Eccentricity e = \(\sqrt {1 + \frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} = \sqrt {1 + \frac{9}{{16}}} = \sqrt {\frac{{16 + 9}}{{16}}} = \sqrt {\frac{{25}}{{16}}} = \frac{5}{4}\)

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 \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is given by e = \(\rm \sqrt{1+\frac{b^2}{a^2}}\), for a > b.

Calculation:

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

Eccentricity is √2.

⇒ √2 = \(\rm \sqrt{1+\frac{b^2}{a^2}}\)

Squaring both sides, we get

⇒ 2 = \(\rm 1+\frac{b^2}{a^2}\)

⇒ \(\rm \frac{b^2}{a^2}\) = 1

⇒ a² = b²

The required equation is therefore, \(\rm \frac{x^2}{a^2}-\frac{y^2}{a^2}=1\) 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

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