- x² = 16 y
- x = 0
- (x + 4)² = 2/3 (y - 2)
- 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.
- (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)
