Question

point
The foci of the hyperbola 9y2 - 4x2 = 36, are:-

Answer 1

EXPLANATION.

Equation of Hyperbola

$\sf \: \implies \: 9 {y}^{2} - 4 {x}^{2} = 36 \\ \\ \sf \: \implies \: \frac{ {9y}^{2} }{36} \: - \: \frac{4 {x}^{2} }{36} = 1 \\ \\ \sf \: \implies \: \frac{ {y}^{2} }{4} \: - \: \frac{ {x}^{2} }{9} = 1 \\ \\ \sf \: \implies \: \frac{ {x}^{2} }{9} \: - \: \frac{ {y}^{2} }{4} = - 1$

$\sf \: \implies(1) = centre \: = (0,0) \\ \\ \sf \: \implies \: vertex \: = (0,b) \: \: and \: (0, - b) \\ \\ \sf \: \implies \: vertex \: = (0,2) \: and \: (0, - 2)$

$\sf \: \implies \:eccentricity = {a}^{2} = {b}^{2}( {e}^{2} - 1) \\ \\ \sf \: \implies \:9 = 4( {e}^{2} - 1) \\ \\\sf \: \implies \:9 = 4 {e}^{2} - 4 \\ \\ \sf \: \implies \:13 = 4 {e}^{2} \\ \\ \sf \: \implies \: \: {e}^{2} = \frac{13}{4} = \frac{ \sqrt{13} }{2}$

$\sf \: \implies \:foci \: = (0,be) \: and \: (0, - be) \\ \\ \sf \: \implies \: foci \: = (0,2 \times \frac{ \sqrt{13} }{2} ) \: and \: (0, - 2 \times \frac{ \sqrt{13} }{2} ) \\ \\ \sf \: \implies \: foci \: = (0, \sqrt{13} ) \: and \: (0, - \sqrt{13})$

$\sf \: \implies \: equation \: of \: directrix \: = y \: = \dfrac{b}{e} and \: y \: = \dfrac{ - b}{e} \\ \\ \sf \: \implies \: equation \: of \: directrix \: = y \: = \frac{2}{ \sqrt{13} } \times 2 \: \: and \: \: y \: = \frac{ - 2}{ \sqrt{13} } \times 2 \\ \\ \sf \: \implies \: equation \: of \: directrix \: = y \: = \frac{4}{ \sqrt{13} } \: \: and \: \: y \: = \frac{ - 4}{ \sqrt{13} }$

$\sf \: \implies \: length \: of \: transverse \: axis \: = 2b \: = 2(2) = 4 \\ \\ \sf \: \implies \: length \: of \: conjugate \: axis \: 2a \: = 2(3) = 6 \\ \\ \sf \: \implies \: length \: of \: latus \: rectum \: = \frac{2 {a}^{2} }{b} = \frac{2(3) {}^{2} }{2} = 9$