1. Find the vertex, focus and directrix for the parabola given by
$y = 2x^2$
$(y-0) = 2(x-0)^2$
$(x-0)^2 = \frac{1}{2}(y-0)$
$4p = \frac{1}{2} \implies p = \frac{1}{8}$
V:$(0,0)$
F: $(0,\frac{1}{8})$
D: $y = -\frac{1}{8}$
2. Find the vertex, focus and directrix for the parabola given by
$$y = -2018x^2$$
$(x-0)^2 = -\frac{1}{2018}(y-0)$
$4p = -\frac{1}{2018} \implies p = -\frac{1}{8072}$
V: $(0,0)$
F: $(0,-\frac{1}{8072})$
D: $y = \frac{1}{8072}$
3. Find the vertex, focus and directrix for the parabola given by
$(y-2)^2 = 8(x+5)$
$4p = 8 \implies p = 2$
V: $(-5,2)$
F: $(-3,2)$
D: $x= -7$
4. Find the vertex, focus and directrix for the parabola given by
$(y+6)^2 = \frac{1}{2}(x-1)$
$4p = \frac{1}{2} \implies p = \frac{1}{8}$
V:$(1,-6)$
F:$(\frac{9}{8},-6)$
D:$x = \frac{7}{8}$
5. Find the vertex, focus and directrix for the parabola given by
$y = 2x^2+5x-7$
$x^2+\frac{5}{2}x-\frac{7}{2} = \frac{y}{2}$
$x^2+\frac{5}{2}x = \frac{y}{2} +\frac{7}{2}$
$x^2+\frac{5}{2}x +\frac{25}{16}= \frac{y}{2} +\frac{7}{2}+\frac{25}{16}$
$(x+\frac{5}{4})^2 = \frac{81}{16}+\frac{y}{2}$
$(x+\frac{5}{4})^2 = \frac{1}{2}(\frac{81}{8}+y)$
$4p = \frac{1}{2} \implies p = \frac{1}{8}$
V:$(-\frac{5}{2}, -\frac{81}{8})$
F:$(-\frac{5}{2}, -\frac{80}{8})$
D:$y = -\frac{82}{8}$
6. Find the vertex, focus and directrix for the parabola given by
$y = -\frac{x^2}{4} – 2x +8$
$-4y = x^2 +8x -32$
$x^2 +8x +16 = 16+32+(-4y)$
$(x+4)^2 = 48 – 4y$
$(x+4)^2 = 4(12 – y)$
$(x+4)^2 = -4(y-12)$
$4p = -4 \implies p= -1$
V:$(-4,12)$
F:$(-4,11)$
D:$y = 13$
7. Find the equation of the parabola which passes through the point $(8,12)$ with a vertex of $(4,-2)$
This one I'm a little confused about, aren't there multiple parabolas which could could through this point and have this vertex? I'm just going to assume it's squared in the x term, so:
$(x-4)^2 = 4p(y+2)$
$(8-4)^2 = 4p(12+2)$
$16 = 4p(14)$
$\frac{16}{14} = 4p \implies p = \frac{4}{14}$
The equation is $(x-4)^2 = \frac{16}{14} (y+2)$
8. Explain, with words, how the distance between the vertex and focus of a
parabola affects the steepness of the parabola
As the distance between the focus and directrix increases, $|p|$ decreases which means the parabola widens.
Best Answer
All the other answers are correct, equation of directrix is wrong in the fourth question (it should be $\frac{7}{8}$), the final answer written in the seventh question is wrong $\Big($putting value of 4p the final equation becomes $(x-4)^2 = \frac{16}{14}(y+2)\Big)$ and in the eighth question, we know that the distance between focus and directrix is 2|p|, therefore, as the distance increases, |p| increases.
Hope it is helpful:)