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Quadrilateral DXHW∼quadrilateral LSCT
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Because the focus is (-2,-2) and the directrix is y = -4, the vertex is (-2,-3).
Consider an arbitrary point (x,y) on the parabola.
The square of the distance between the focus and P is
(y+2)² + (x+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:
Consider an arbitrary point (x,y) on the parabola.
The square of the distance between the focus and P is
(y+2)² + (x+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: