Question

find the vertex, focus, and directrix for the following parabolas.

(a)
\\((y - 9)^2 = 16(x - 4)\\)

vertex :

focus :

directrix:

(b)
\\(y^2 - 8y = 8x - 4^2\\)

vertex :

focus :

directrix

Explanation:

Step1: Identify standard form for (a)

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

Step2: Extract vertex for (a)

$$(h, k) = (4, 9)$$

Step3: Calculate value of p for (a)

$$4p = 16 \implies p = 4$$

Step4: Find focus for (a)

$$(h + p, k) = (4 + 4, 9) = (8, 9)$$

Step5: Find directrix for (a)

$$x = h - p \implies x = 4 - 4 = 0$$

Step6: Rewrite equation (b) by completing square

$$y^2 - 8y + 16 = 8x - 16 + 16 \implies (y - 4)^2 = 8x$$

Step7: Extract vertex for (b)

$$(h, k) = (0, 4)$$

(Note: Correcting the equation simplification: \(y^2 - 8y = 8x - 16 \implies y^2 - 8y + 16 = 8x \implies (y - 4)^2 = 8(x - 0)\). Thus, the vertex is \((0, 4)\).)

Step8: Recalculate vertex for (b)

$$(h, k) = (0, 4)$$

Step9: Calculate value of p for (b)

$$4p = 8 \implies p = 2$$

Step10: Find focus for (b)

$$(h + p, k) = (0 + 2, 4) = (2, 4)$$

Step11: Find directrix for (b)

$$x = h - p \implies x = 0 - 2 = -2$$

Answer:

(a)
vertex: \((4, 9)\)
focus: \((8, 9)\)
directrix: \(x = 0\)

(b)
vertex: \((4, 4)\)
focus: \((6, 4)\)
directrix: \(x = 2\)