The foci of the hyperbola 2{x^2} - 3{y^2} = 5 | vedclass

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Question

(a) The given equation is $2{x^2} - 3{y^2} = 5$

$ \Rightarrow \,$$\frac{{{x^2}}}{{5/2}} - \frac{{{y^2}}}{{5/3}} = 1$

Now ${b^2} = {a^2}({e^2} -

Vedclass · Accepted Answer

$\left( { \pm \frac{5}{{\sqrt 6 }},\;0} \right)$

Vedclass · Answer

(a) The given equation is $2{x^2} - 3{y^2} = 5$

$ \Rightarrow \,$$\frac{{{x^2}}}{{5/2}} - \frac{{{y^2}}}{{5/3}} = 1$

Now ${b^2} = {a^2}({e^2} - 1)$

$ \Rightarrow \,\frac{5}{3} = \frac{5}{2}({e^2} - 1)$

==> $e = \sqrt {\frac{5}{3}} $.

The foci of hyperbola $( \pm \,ae,\,0)$

$ = \left( { \pm \sqrt {\frac{5}{2}} \,.\,\sqrt {\frac{5}{3}} ,0} \right)$ $ = \left( { \pm \,\frac{5}{{\sqrt 6 }},0} \right)$.

The foci of the hyperbola $2{x^2} - 3{y^2} = 5$, is

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