Module 7 project.
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Refer to the diagram shown below.
The directrix is y = -4 and the focus is (-2, -2).
Therefore the vertex is at (-2, -3).
Consider an arbitrary point (x,y) on the parabola.
The square of distance from the focus to the point is
(x+2)² + (y+2)²
The square of the distance from the point to the directrix is
(y+4)²
Therefore
(y+4)² = (y+2)² + (x+2)²
y² + 8y + 16 = y² + 4y + 4 + (x+2)²
4y = (x+2)² - 12
y = (1/4)(x+2)² - 3
Answer:
The directrix is y = -4 and the focus is (-2, -2).
Therefore the vertex is at (-2, -3).
Consider an arbitrary point (x,y) on the parabola.
The square of distance from the focus to the point is
(x+2)² + (y+2)²
The square of the distance from the point to the directrix is
(y+4)²
Therefore
(y+4)² = (y+2)² + (x+2)²
y² + 8y + 16 = y² + 4y + 4 + (x+2)²
4y = (x+2)² - 12
y = (1/4)(x+2)² - 3
Answer: