Refer to the diagram shown.
At an arbitrary point P (x,y) on the parabola, the distance from the focus should be equal to the distance from the directrix.
The distance of P from the focus is
d₁ = √[x² + (y+3)²]
The distance from P to the directrix is
d₂ = 3 - y
Therefore
d₁ = d₂.
That is,
x² + (y+3)² = (3 - y)²
x² + y² + 6y + 9 = 9 - 6y + y²
x² + 12y = 0
y = - (1/12) x²
Answer:
At an arbitrary point P (x,y) on the parabola, the distance from the focus should be equal to the distance from the directrix.
The distance of P from the focus is
d₁ = √[x² + (y+3)²]
The distance from P to the directrix is
d₂ = 3 - y
Therefore
d₁ = d₂.
That is,
x² + (y+3)² = (3 - y)²
x² + y² + 6y + 9 = 9 - 6y + y²
x² + 12y = 0
y = - (1/12) x²
Answer: