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Let 1 be the directrix of the parabola $9 y^2+12 y+9 x-14=0$ and $l_1$ be the line passing through the vertex of this parabola and the origin. If $(\mathrm{h}, \mathrm{k})$ is the point of intersection of $l$ and $l_1$, then $\mathrm{h}+\mathrm{k}=$
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The correct answer is:
$-9 / 2$
Given: $9 y^2+12 y+9 x-14=0$
$\Rightarrow\left(y+\frac{2}{3}\right)^2=-4 \times \frac{1}{4}(x-2)$
Whose vertex is $\left(-\frac{2}{3}, 2\right)$ Equation of line passing through $\left(-\frac{2}{3}, 2\right)$ and $(0,0)$ :
$(y-0)=\frac{2-0}{-\frac{2}{3}-0}(x-0) \Rightarrow y=-3 x$
Now, the equation of directrix of parabola (i) is
$x=\frac{1}{4}+2 \Rightarrow x=\frac{9}{4}$ ...(iii)
Solving $\mathrm{eq}^{\mathrm{n}}$ (ii) \& (iii), we get
$x=\frac{9}{4}$ and $y=-\frac{27}{4} \quad \therefore h=\frac{9}{4} \& k=-\frac{27}{4}$
So, $h+k=\frac{9}{4}-\frac{27}{4}=\frac{-18}{4}=\frac{-9}{2}$
$\Rightarrow\left(y+\frac{2}{3}\right)^2=-4 \times \frac{1}{4}(x-2)$
Whose vertex is $\left(-\frac{2}{3}, 2\right)$ Equation of line passing through $\left(-\frac{2}{3}, 2\right)$ and $(0,0)$ :
$(y-0)=\frac{2-0}{-\frac{2}{3}-0}(x-0) \Rightarrow y=-3 x$
Now, the equation of directrix of parabola (i) is
$x=\frac{1}{4}+2 \Rightarrow x=\frac{9}{4}$ ...(iii)
Solving $\mathrm{eq}^{\mathrm{n}}$ (ii) \& (iii), we get
$x=\frac{9}{4}$ and $y=-\frac{27}{4} \quad \therefore h=\frac{9}{4} \& k=-\frac{27}{4}$
So, $h+k=\frac{9}{4}-\frac{27}{4}=\frac{-18}{4}=\frac{-9}{2}$
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