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If ` 5x +9 =0 ` is the directtrix of the hyperbola `16x^(2) -9y^(2)=144,` then its corresponding focus isA. `(-(5)/(3),0)`B. `(-5,0)`C. `((5)/(3),0)`D. `(5,0)` |
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Answer» Equation of given hyperbola is `16x^(2)-9y^(2)=144` `rArr (x^(2))/(9)-(y^(2))/(16)=1 " ...(i)" ` So, the eccentricity of Eq. (i) `e=sqrt(1+(16)/(9))=(5)/(3)` [` because ` the eccentricity (e) of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is `sqrt(1+(b//a)^(2))`] and given directrix is `5x+9=0 rArr x= -9//5` So, corresponding focus is `(-3((5)/(3)),0)=(-5,0)` |
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