Question

16. The eccentricity of the conjugate hyperbola
of the hyperbola x^2- 3y^2=1 is
a. 2 b. 2'13 c. 4 d. 4/3​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Eccentricity=\sqrt{\frac{5}{3}}}}}$

$\green{\tt{\therefore{Eccentricity=\frac{2}{\sqrt{3}}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }}\\ \tt: \implies Eqn \: of \: hyperbola = {x}^{2} - 3 {y}^{2} = 1 \\ \\ \red{ \underline \bold{To \: Find: }} \\ \tt: \implies Eccentricity = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies Eqn \: of \: hyperbola = {x}^{2} - 3 {y}^{2} = 1 \\ \\ \tt: \implies Conjugate \: hyperbola = {x}^{2} - 3 {y}^{2} = - 1 \\ \\ \tt: \implies {x}^{2} - 3 {y}^{2} = - 1 \\ \\ \tt: \implies \frac{ {x}^{2} }{1} - \frac{ {y}^{2} }{ \frac{1}{3} } = - 1 \\ \\ \bold{Where : } \\ \tt: \implies {a}^{2} = 1 \\ \\ \tt: \implies {b}^{2} = \frac{1}{3} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {a}^{2} = {b}^{2} ( {e}^{2} - 1) \\ \\ \tt: \implies 1 = \frac{1}{3} ( {e}^{2} - 1) \\ \\ \tt: \implies 1 - \frac{1}{3} = {e}^{2} - 1 \\ \\ \tt: \implies \frac{2}{3} = {e}^{2} - 1$

$\tt: \implies \frac{2}{3} + 1 = {e}^{2} \\ \\ \tt: \implies \frac{5}{3} = {e}^{2} \\ \\ \green{\tt: \implies e = \sqrt{ \frac{5}{3} } } \\ \\ \bold{Eccentricity \: of \: standard \: hyperbola} \\ \tt: \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1) \\ \\ \tt: \implies \frac{1}{3} = 1( {e}^{2} - 1) \\ \\ \tt: \implies \frac{1}{3} = {e}^{2} - 1 \\ \\ \tt: \implies \frac{1}{3} + 1 = {e}^{2} \\ \\ \tt: \implies \frac{4}{3} = {e}^{2} \\ \\ \tt: \implies e = \sqrt{ \frac{4}{3} } \\ \\ \green{\tt: \implies e = \frac{2}{ \sqrt{3} } }$

Answer 2

$\tt{\huge{\orange{----------}}} B.Q$

QUESTION :

16. The eccentricity of the conjugate hyperbola

16. The eccentricity of the conjugate hyperbolaof the hyperbola x^2- 3y^2=1 is ..........

a. 2

b. 2/13

c. 4

d. 4/3

SOLUTION :

First we will have to find the Equation of the conjugate Hyperbola.

Equation of Original Hyperbola :

X ^ 2 - 3 Y ^2 = 1

Equation of the conjugate Hyperbola can be found by multiplying the original Equation by ( - 1 )

Hence Equation Of Conjugate Hyperbola :

X ^ 2 - 3 Y ^2 = -1

Breaking this into Intercept Form :::

=> 3 can be written as : 1 / { 1 / 3 }

=> - 1 can be written as : 1 / { -1 }

Hence Equation of Conjugate Hyperbola In Intercept Form :

=> X ^ 2 / { -1 } + Y ^ 2 / { 1 / 3 } = 1

So, X Intercept A = -1

Y Intercept B = 1 / 3

for Standaard Hyperbola :

=> { A } ^ 2 = 1

=> { B } ^ 2 = { 1 / 3 }

Now we know that :

b^2 =a ^ 2 ( e ^2 - 1 )

=> { 1 / 3 } = e^2 - 1

=> e^2 = 1 +{ 1 / 3 } = 4 / 3

=>e = + { [ 2 ]/√ [ 3 ] } and - { [ 2 ] / [√ 3 ] }......[ A ]

similarly solving for conjugate Hyperbola,

e = + 5 / √ { 3 } and - 5 / √ { 3 }

N o N e O f T h E g I V e N O p T I O N s A r E c O R R e C T.