Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola with the given equation. Then find the length of the latus rectum and graph the parabola.
$x=-4 y^2+6 y+2$

Solution

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Rewrite in standard form. $$\begin{aligned} x&=-4y^2+6y+2\\ x&=-4\left[y^2-\frac{6}{4}+\left(\frac{b}{2}\right)^2\right]+2-(-4)\left(\frac{b}{2}\right)^2\\ x&=-4\left[y^2-1.5+\left(\frac{1.5}{2}\right)^2\right]+2+4\left(\frac{1.5}{2}\right)^2\\ x&=-4\left[y^2-1.5+(0.75)^2\right]+2+4(0.75)^2\\ x&=-4\left(y^2-1.5+(0.5625\right)+2+4(0.5625)\\ x&=-4(y-0.75)^2+4.25 \end{aligned}$$ Thus, the standard form is $$\begin{aligned} x&=-4(y-0.75)^2+4.25 \end{aligned}$$ Therefore, $\text{vertex}=(4.25,0.75)$ $\text{axis of symmetry}=0.75$ $$\begin{aligned}\text{focus}&=\left (h+\dfrac{1}{4a},k\right)\\ &=\left (4.25+\frac{1}{4(-4)},0.75\right)\\ &=\left (4.25-\frac{1}{16},0.75\right)\\ &=\left (\frac{68-1}{16},-0.75\right)\\ &=\left (\frac{69}{16},-0.75\right)\\ \end{aligned}$$ $\text{focus}=\left (4.1875,0.75\right)\\$ $$\begin{aligned}\text{directrix}:x&=h-\dfrac{1}{4a}\\ &=4.25-\dfrac{1}{4(-4)}\\ &=4.25+\left(\frac{1}{16}\right)\\ &=\frac{68+1}{16}\\ \end{aligned}$$ $\text{directrix}=4.3125$ $a<0= \text {opens to the left}$ $$\begin{aligned}\text{lenght of latus rectum}&=\dfrac{1}{a}\text{ unit}\\&=\dfrac{1}{-4} \end{aligned}$$ $\text{latus rectum}=\frac{1}{4} \text{ unit}$