Find the coordinate of the foci, vertice eccentricity and the length of the latus rectum of the hyperbola
`49y ^(2) - 16x ^(2) = 784`

Explanation

Equation of hyperbola
`49y^(2)-16x^(2)=784`
`rArr" "(y^(2))/(16)-(x^(2))/(49)=1`
The transverse axis of hyperbola is along y-axis.
Comparing with `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1`
`b^(2)=16," "a^(2)=49`
`rArrb=4," "a=7`
`:. "Vertices"-=(0,pmb)-=(0,pm4)`
Eccentricity `e=sqrt(1+(a^(2))/(b^(2)))`
`=sqrt(1+(49)/(16))=sqrt((65)/16)=sqrt((65)/4)`
Now, `be4xx=(sqrt(65))/(4)=sqrt(65)`
`:. "Coordinates of foci"-=(0,pmbe)-=(0,pmsqrt(65))`
Length of latus latus rectum `=(2a^(2))/(b)=(2xx49)/(4)=(49)/(2)`