Question

the eccentricity of the hyperbola
$4{x }^{2} - 9 {y}^{2} = 2ax + {b}^{2}$
​

Answer 1

Answer:

$\sf{e=\dfrac{\sqrt{13}}{3}}$

Step-by-step explanation:

$\sf{\implies4x^2-9y^2=2ax+b^2}\\\sf{\implies4x^2-2ax-9y^2=b^2}\\\sf{\implies4\left(x^2+\dfrac{ax}{2}\right)-9y^2=b^2}\\\sf{\implies4\left(x^2-\dfrac{2ax}{4}+\dfrac{a^2}{16}-9y^2\right)=b^2-\dfrac{a^2}{4}}\\\sf{\implies4\left(x-\dfrac{a}{4}\right)^2-9y^2=b^2-\dfrac{a^2}{4}=k(say)}\\\sf{\implies\dfrac{\left(x-\dfrac{a}{4}\right)^2}{\dfrac{k}{4}}-\dfrac{y^2}{\dfrac{k}{9}}}\\\sf{\implies\:Let\:k>0}\\\sf{\implies\:Therefore,\:e=\sqrt{1+\dfrac{\dfrac{k}{9}}{\dfrac{k}{4}}}}\\\sf{\implies\:e=\sqrt{1+\dfrac{4}{9}}}\\\\\boxed{\sf{e=\dfrac{\sqrt{13}}{3}}}$

Answer 2

$\huge\underline \mathbb {SOLUTION:-}$

Answer:

• $\mathsf \red {e = \frac{ \sqrt{13} }{3} }$

Given:

• $\mathsf \green {4x {}^{2} - 9y {}^{2} = 2ax + b {}^{2} }$

$\underline \mathsf \pink {According\:to\:question\: :-}$

★ $\mathsf \green {4x {}^{2} - 9y {}^{2} = 2ax + b {}^{2} }$

$\leadsto \mathsf {4x {}^{2} - 2ax - 9y {}^{2} = b {}^{2} }$

$\leadsto \mathsf {4(x {}^{2} + \frac{ax}{2} ) - 9y {}^{2} = b {}^{2}}$

$\leadsto \mathsf {4(x {}^{2} - \frac{2ax}{4} + \frac{a {}^{2} }{16} - 9y {}^{2}) = b^{2} - \frac{a^{2}}{4} }$

$\leadsto \mathsf {4(x - \frac{a}{4} ) {}^{2} - 9y {}^{2} = b {}^{2} - \frac{a {}^{2} }{4} = k \: (Say)}$

$\leadsto$$\frac{(x - \frac{a}{4}) {}^{2} }{ \frac{k}{4} } - \frac{y {}^{2} }{ \frac{k}{9} }$

$\underline \mathsf \blue {Let\::-}$

• $\mathsf {k > 0}$

$\underline \mathsf \pink {Hence\: :-}$

$\leadsto$$e = \sqrt{1 + \frac{ \frac{k}{9} }{ \frac{k}{4} } }$

$\leadsto$$e = \sqrt{1 + \frac{4}{9} }$

$\mathsf \red {e = \frac{ \sqrt{13} }{3} }$

$\underline \mathsf \blue {Therefore \: :-}$

• $\mathsf \red {e = \frac{ \sqrt{13} }{3} }$

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