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Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the Hyperbola `49y^(2)-16x^(2)=784` |
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Answer» 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)` |
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