The latus rectum of the conic `3x^2+4y^2-6x+8y-5=0` is a.`3` b. `(sqrt(3))/2` c. `2/(sqrt(3))` d. none of these

Explanation

`3 x^{2}+4 y^{2}-6 x+8 y-5=0`

`Rightarrow 3(x^{2}-2 x)+4(y^{2}+2 y)=5`

`Rightarrow 3(x^{2}-2 x+1)+4(y^{2}+2 y+1)=5+3+4`

`Rightarrow 3(x-1)^{2}+4(y+1)^{2}=12`

`Rightarrow frac{(x-1)^{2}}{4}+frac{(y+1)^{2}}{3}=1`

So, `a=2` and `b=sqrt{3}`

`therefore` Latus rectum `=frac{2 b^{2}}{a}`

`=(2 frac{[sqrt{3}]^{2}}{2})`

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