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Find all the aspects of hyperbola `16x^(2)-3y^(2)-32x+12y-44=0.` |
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Answer» Correct Answer - `e=sqrt(19//3),"Vertices"-=(1pmsqrt3,2),"Foci"-=(1pmsqrt(19),2), "Directrix:"x-1=pm3//sqrt(19)` We have, `16(x^(2)-2x)-3(y^(2)-4y)=44` `"or "16(x-1)^(2)-3(y-2)^(2)=48` `"or "((x-1)^(2))/(3)-((y-2)^(2))/(16)=1` Clearly, transverse axis is horizontal, having equation `y-2=0.` Conjugate axis is `x-1=0 or x=1.` Centre is (1,2). Here, `a=sqrt3` and b = 4. Thus, length of transverse axis is `2sqrt3` and that of conjugate axis is 8. `"Eccentricity, e"=sqrt(1+(16)/(3))=sqrt((19)/(3))` Vertices lie at distance a from centre on transverse axis. Therefore, vertices are `(1pm sqrt3, 2)` Foci lie at distance ae from centre on transverse axis. Therefore, foci are `(1pm sqrt(19),2)`. Equation of directrices are `x-1= pm(1)/(e)` `"or "x-1=pm(3)/(sqrt(19)` |
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