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જેથી નાભિઓ $(6, 5), (-4, 5)$ હોય અને ઉત્કેન્દ્રતા $5/4$ હોય તેવા અતિવલયનું સમીકરણ :
$\frac{{{{\left( {x\,\, - \,\,1} \right)}^2}}}{{16}}\,\, - \,\,\frac{{{{\left( {y\, - \,\,5} \right)}^2}}}{9}\,\, = \,1$
$\frac{{{x^2}}}{{16}}\,\, - \,\,\frac{{{y^2}}}{9}\,\, = \,\,1$
$\frac{{{{\left( {x\,\, - \,\,1} \right)}^2}}}{{16}}\,\, - \,\,\frac{{{{\left( {y\,\, - \,\,5} \right)}^2}}}{9}\,\, = \,\, - 1$
આપેલ પૈકી એક પણ નહિ
Solution
Let the centre of hyperbola be $(\alpha, \beta)$
As $y=5$ line has the foci, it also has the major axis.
$\therefore \frac{(x-\alpha)^2}{a^2}-\frac{(y-\beta)^2}{b^2}=1$
Midpoint of foci $=$ centre of hyperbola
$\therefore \alpha=1, \beta=5$
Given, $e=\frac{5}{4}$.
We know that foci is given by $(\alpha=a, \beta)$
$\therefore \alpha+a a =6$
$\Rightarrow 1+\frac{5}{4} a =6 \Rightarrow a =4$
Using $b^2=a^2\left(e^2-1\right)$
$\Rightarrow b^2=16\left(\frac{25}{16}-1\right)=9$
$\therefore$ Equation of hyperbola $\Rightarrow \frac{( x -1)^2}{16}-\frac{( y -5)^2}{9}=1$
