Let l be the directrix of the parabola 9y^2+1 | vedclass

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(B) Given equation: $9y^2+12y+9x-14=0$ Rewrite as: $9(y^2 + \frac{4}{3}y) = -9x + 14$ $9(y + \frac{2}{3})^2 = -9x + 14 + 4 = -9x + 18$ $(y + \frac{2}{

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$3/2$

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(B) Given equation: $9y^2+12y+9x-14=0$ Rewrite as: $9(y^2 + \frac{4}{3}y) = -9x + 14$ $9(y + \frac{2}{3})^2 = -9x + 14 + 4 = -9x + 18$ $(y + \frac{2}{3})^2 = -(x - 2)$ Comparing with $(y-k')^2 = -4a(x-h')$,we get vertex $(h', k') = (2, -2/3)$ and $4a = 1 \Rightarrow a = 1/4$. The directrix $l$ is $x = h' + a = 2 + 1/4 = 9/4$. The line $l_1$ passes through $(2, -2/3)$ and $(0, 0)$,so its slope is $m = \frac{-2/3 - 0}{2 - 0} = -1/3$. Equation of $l_1$: $y = -\frac{1}{3}x$. Intersection of $l$ $(x = 9/4)$ and $l_1$ $(y = -x/3)$: $h = 9/4$,$k = -\frac{1}{3}(9/4) = -3/4$. Thus,$h+k = 9/4 - 3/4 = 6/4 = 3/2$.

Let $l$ be the directrix of the parabola $9y^2+12y+9x-14=0$ and $l_1$ be the line passing through the vertex of this parabola and the origin. If $(h, k)$ is the point of intersection of $l$ and $l_1$,then $h+k=$

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