Parabola: Problems with Solutions

Solution:
$(y+4)^{2}=4\left( x+1\right)$
When $x=0$ then $(y+4)^{2}=4$

$y+4=\pm 2\Longrightarrow y=-2$ and $y=-6$ are $y-axis$ intersection.

When $y=0$, $4\left( x+1\right) =16\Longrightarrow x=3$

Answer: $(0,-2); (0,-6); (3,0);$

Solution:
Since the focus is left the directrix the equation is

$y^{2}=4px$ and $p=-2$

So $y^{2}=-8x$ is the equation of the parabola.

Solution:
Since the focus is on the left of the directrix the equation is of the form
$\left( y-k\right) =4p(x-h)^{2}$, $p=-3$ and $(h,k)=(-3,-1)$

So
$\left( y+1\right) =-12(x+3)^{2}$ is the equation of the parabola

and $p=-3$ then focus is $(-3,-1+p)=(-3,-4)$

Solution:
Answer: Vertex $(-4,2)$, focus $(-\frac{127}{32},2)$, axis $y=2$, directrix $x=-\frac{1}{32}$

We will complete the square.
$y^{2}-4y-8x-28=(y-2)^{2}-4=8x+28\Longrightarrow (y-2)^{2}=8x+32$

So the equation of the parabola is $\frac{1}{8}(y-2)^{2}=(x+4)$

The vertex is at $(-4,2)$, $p=\frac{1}{32}$ The focus is $(-4+p,2)=(-\frac{127}{32},2)$,
the axis is $y=2$ and directrix is $x=-p=-\frac{1}{32}$

Solution:
Answer: vertex $(0,1)$, focus $(4,1)$, directrix $x=-4$, axis $y=1$

This parabola has the form $\left(y-k\right)^{2}=4p(x-h)$

$(h,k)=(0,1)$, $p=4$ then vertex is at $(0,1)$, focus at $(0+p,1)=(4,1)$

directrix $x=h-p\Longrightarrow x=-4$

The axis is $y=k\Longrightarrow y=1$

Solution:
Answer: Vertex: $(−5,−1)$, focus: $(−5,−2)$, axis: $x=-5$, directrix: $y=0$

This parabola is of the form $\left( x-h\right)^{2}=4p(y-k)$, $(h,k)=(-5,-1)$
$p=-1$, vertex: $(-5,-1)$, focus: $(-5,-1+p)=(-5,-2)$

Directrix $y=k-p\Longrightarrow y=0$

Axis $x=h\Longrightarrow x=-5$

Solution:
Answer: $x^{2}=28y$

This parabola have vertex at the midpoint between the focus and directrix. The vertex is $(0,\frac{7-7}{2})=(0,0)$ and $p=7$

So the equation is of the form $\left( x-h\right)^{2}=4p(y-k)$

Then $\left(x-0\right)^{2}=28(y-0)\Longrightarrow x^{2}=28y$

Solution:
Answer: $\left(x-5\right)^{2}=-24(y-1)$

In this case $p=-\left\vert 7-1\right\vert =-6$ and $(h,k)=(5,1)$
The equation is of the form $\left(x-h\right)^{2}=4p(y-k)$

So $\left( x-5\right)^{2}=-24(y-1)$

Solution:
Answer: $x^{2}=\frac{1}{2}y$

The equation of the parabola which axis coinsides with the y-axis is of the form
$\left( x-h\right)^{2}=4p(y-k)$

Now if the vertex is $(0,0)=(h,k)$ we get
$\left( x-0\right) ^{2}=4p(y-0)\Longrightarrow x^{2}=4py$ and we have to find $c$

Since the parabola is through point $(-2,8)$
$(-2)^{2}=4p\cdot 8=32c\Longrightarrow p=\frac{1}{8}$ so
$x^{2}=\frac{1}{2}y$ is the equation of the parabola.