Question

Find the principal axis, vertex, focus, directrix, endpoints of the focal width and length of the focal width, and sketch the graph of the following:

1. -(x + 5)^2 = y

2. (y - 8)^2 = 24(x + 1)

3. 4x - y^2 = 0

Answer 1

Answer on Question #62579 – Math – Analytic Geometry

Question

Find the principal axis, vertex, focus, directrix, endpoints of the focal width and length of the focal width, and sketch the graph of the following

1. $-(x + 5)^2 = y$

Solution

1. Standard form of equation for a vertical parabola

$$(x - h)^2 = 4p(y - k)$$

$(h,k)$ being the $(x,y)$ coordinates of the vertex; principal axis: $x - h = 0$; focus: $F(h,p + k)$; directrix: $y = -p + k$ (directrix is horizontal); length of the focal width: $d = 4|p|$; endpoints of the focal width: $B(h - 2p, p + k), C(h + 2p, p + k)$.

For given equation $-(x + 5)^2 = y$, or $(x + 5)^2 = -y$ we have: $h = -5, k = 0, p = -\frac{1}{4}$.

If $p = -\frac{1}{4}$, then parabola opens down. So

principal axis: $x = -5$;

vertex: $A(-5,0)$;

focus: $F\left(-5, -\frac{1}{4}\right)$;

directrix: $y = \frac{1}{4}$;

endpoints of the focal width: $B\left(-5\frac{1}{2}, -\frac{1}{4}\right)$, $C\left(-4\frac{1}{2}, -\frac{1}{4}\right)$;

length of the focal width: $d = 1$.

Sketch the graph:

Answer:

principal axis: $x = -5$;

vertex: $A(-5,0)$;

focus: $\mathrm{F}\left( {-5, - \frac{1}{4}}\right)$ ;

directrix: $y = \frac{1}{4}$

endpoints of the focal width: $B\left(-5\frac{1}{2}, - \frac{1}{4}\right),C\left(-4\frac{1}{2}, - \frac{1}{4}\right);$

length of the focal width: $d = 1$ .

Question

Find the principal axis, vertex, focus, directrix, endpoints of the focal width and length of the focal width, and sketch the graph of the following

2. $(y - 8)^{2} = 24x + 1$

Solution

2. Standard form of equation for a horizontal parabola

$$(y - k) ^ {2} = 4 p (x - h)$$

where $(h,k)$ is the coordinates of the vertex and $p$ is the distance from the vertex to the focus. Principal axis: $y - k = 0$ ; focus: $F(p + h,k)$ ; directrix: $x = -p + h$ (directrix is vertical); length of the focal width: $d = 4|p|$ ;

endpoints of the focal width: $B(p + h, k + 2p)$ , $C(p + h, k - 2p)$ .

For given equation $(y - 8)^2 = 24x + 1$ , or $(y - 8)^2 = 24\left(x + \frac{1}{24}\right)$ we have $h = -\frac{1}{24}$ , $k = 8$ , $p = 6$ . If $p = 6 > 0$ , then parabola opens right. So

principal axis: $y = 8$

vertex: $A\left(-\frac{1}{24},8\right)$

focus: $F\left(5\frac{23}{24},8\right)$

directrix: $x = -6\frac{1}{24}$

endpoints of the focal width: $B\left(5\frac{23}{24}, 20\right)$ , $C\left(5\frac{23}{24}, -4\right)$ ;

length of the focal width: $d = 24$ .

Sketch the graph:

Answer:

principal axis: $y = 8$

vertex: $A\left(-\frac{1}{24},8\right)$

focus: $F\left( {5\frac{23}{24},8}\right)$ ;

directrix: $x = -6\frac{1}{24}$

endpoints of the focal width: $B\left( {5\frac{23}{24},20}\right) ,C\left( {5\frac{23}{24}, - 4}\right)$ ;

length of the focal width: $d = 24$ .

Question

Find the principal axis, vertex, focus, directrix, endpoints of the focal width and length of the focal width, and sketch the graph of the following

3. $4x - y^{2} = 0$

Solution

3. For a horizontal parabola $4x - y^{2} = 0$ , or $y^{2} = 4x$ we have: $h = 0$ , $k = 0$ , $p = 1$ .

Then:

principal axis: $y = 0$

vertex: $A(0,0)$

focus: $F(1,0)$

directrix: $x = -1$ (directrix is vertical);

endpoints of the focal width: $B(1,2)$ , $C(1, - 2)$

length of the focal width: $d = 4$

Sketch the graph:

Answer:

principal axis: $y = 0$;

vertex: $A(0,0)$;

focus: $F(1,0)$;

directrix: $x = -1$;

endpoints of the focal width: $B(1,2)$, $C(1,-2)$;

length of the focal width: $d = 4$.

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