Write the length o the latus rectum of the hyperbola `16 x^2-9y^2=144.

1.	Write the length o the latus rectum of the hyperbola `16 x^2-9y^2=144.`
Answer» Correct Answer - `e=5//4, "Vertices"-=(0, pm4), "Foci"-=(0, pm5), L.R. = 9//2, "Directrix;y"=pm16//5` We have hyperbola `16x^(2)-9y^(2)=-144` `"or "(x^(2))/(9)-(y^(2))/(16)=-1` This equation is of the form `(x^(2))/(a^(2))-(y^(2))/(16)=-1`. Hence, x-axis is the conjugate axis and y-axis is the tranverse axis. Now, `a^(2)=9,b^(2)=16." So, "a=3,b=4.` Length of transverse axis = 2b = 8 Length of conjugate axis = 2a = 6 Eccentricity, `e=sqrt(1+(a^(2))/(b^(2)))=sqrt(1+(9)/(16))=(5)/(4)` Vertics are `(0 pmb) or (0, pm4)`. Foci are `(0, pm be) or (0, pm 5)` Length of latus rectum `=(2a^(2))/(b)=(2(3)^(2))/(4)=(9)/(2)` Equation of directrices are `y=pm(b)/(e)` `"or "y=pm(4)/((5//4))` `"or "y=pm(16)/(5)`

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Write the length o the latus rectum of the hyperbola `16 x^2-9y^2=144.`

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