Question

Parabolas. Please help me

Answer 1

1. x² = 16 y
2. x = 0
3. (x + 4)² = 2/3 (y - 2)
4. Gayle identifies that the vertex of the parabola is (3, -1) . The parabola opens right, and the focus is 3 units away from the vertex. The directrix is 6 units from the focus. The focus is the point (6, -1). The directrix of the equation is x = 0.
5. (y - 6)² = 4p (x - 3)

Step-by-step explanation:

1. First figure

We plot the parabola as given in the attached diagram.

As it is facing upwards, the equation goes as x² = 4py

where, p = 4 (refer the attached diagram)

x² = 4py

x² = 4 (4) y

∴, standard form of parabola is x² = 16 y

2. Second figure

(y + 3)² = 4 (x - 1)

Comparing the given equation with the standard form

(y - k)² = 4p (x - h)

Now from this equation we get to know that

h = 1

p = 1

Directrix is x = (h - p)

So, x = 0

3. Third figure

3x² + 24x - 2y + 52 = 0

3x² + 24x = 2y - 52

3 (x² + 8x) = 2 (y - 26)

(x² + 8x) = 2 (y - 26) / 3

Adding 16 on both sides,

x² + 8x + 16 = 2 (y - 26) / 3 + 16

(x + 4)² = 2/3 y - 52/3 + 16

(x + 4)² = 2/3 y - 4/3

(x + 4)² = 2/3 (y - 2)

4. Fourth figure

(y + 1)² = 12 (x - 3)

Comparing the given equation with the standard form

(y - k)² = 4p (x - h)

Now from this equation we get to know that

k = -1

h = 3

p = 3

Gayle identifies that the vertex of the parabola is (3, -1) . The parabola opens right, and the focus is 3 units away from the vertex. The directrix is 6 units from the focus. The focus is the point (6, -1). The directrix of the equation is x = 0.

5. Fifth figure

Focus = (2, 6)

Directrix is x = 4

Therefore, it follows the standard form

(y - k)² = 4p (x - h)

Directrix is given by x = h-p = 4

Focus is given by (h + p, k) = (2, 6)

Solving for (h - p) = 4, (h + p) = 2

2 - p - p = 4

-2p = 2

p = -1

Hence, h = 3

Therefore, the standard form can be written as

(y - 6)² = 4p (x - 3)