If the directrix of the parabola x^2+4y-6x+la | vedclass

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(C) The given equation is $x^2-6x+4y+\lambda=0$. Completing the square for $x$: $(x-3)^2-9+4y+\lambda=0$,which simplifies to $(x-3)^2 = -4y + (9-\lamb

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focus is $(3,-3)$

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(C) The given equation is $x^2-6x+4y+\lambda=0$. Completing the square for $x$: $(x-3)^2-9+4y+\lambda=0$,which simplifies to $(x-3)^2 = -4y + (9-\lambda) = -4(y - \frac{9-\lambda}{4})$. Comparing this with the standard form $(x-h)^2 = -4a(y-k)$,we have $h=3$,$k=\frac{9-\lambda}{4}$,and $a=1$. The directrix of this parabola is $y = k+a$. Given the directrix is $y+1=0$,i.e.,$y=-1$,we equate $k+a = -1$. Substituting $k$ and $a$: $\frac{9-\lambda}{4} + 1 = -1$. $\frac{9-\lambda}{4} = -2 \implies 9-\lambda = -8 \implies \lambda = 17$. However,checking the options,if $\lambda=17$,then $k = \frac{9-17}{4} = -2$. The focus is $(h, k-a) = (3, -2-1) = (3, -3)$. Thus,option $C$ is correct.

If the directrix of the parabola $x^2+4y-6x+\lambda=0$ is $y+1=0$,then which of the following is correct?

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