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The correct match from column-I to column-II in the following is

Column-I
Equation of Hyperbola

Column-II
Foci, eccentric
A $9 x^{2} - 16 y^{2} = 36$ p $(\pm 4, 0), 2$
B $3 x^{2} - y^{2} = 12$ q $(\pm 5, 0), \frac{5}{3}$
C $16 x^{2} - 9 y^{2} = 144$ r $(0, \pm 3), 3 / 2$
D $4 x^{2} - 5 y^{2} + 20 = 0$ s $(\pm \frac{5}{2}, 0), 5 / 4$

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$A \to r, B \to q, c \to p, D \to s$

$A \to s, B \to p, c \to q, D \to r$

$A \to p, B \to s, c \to r, D \to q$

$A \to s, B \to s, c \to p, D \to q$

answer is B.

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Detailed Solution

$\frac{x^{2}}{4} - \frac{y^{2}}{\frac{9}{4}} = 1$
Here, a>b
$∴ e = \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} = \sqrt{\frac{4 + \frac{9}{4}}{4}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$
Focus $= (\pm a e, 0) = (\pm 5, 0)$
Similarly B,C,D;
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