Find the equation of a parabola with focus (3, 4) and directrix y = 1.
2 answers:
Refer to the diagram shown below.
At an arbitrary point P (x,y) on the parabola, the distance, d, from the focus to P should be equal to the distance from P to the directrix.
That is,
d² = (x - 3)² + (y - 4)² = (y - 1)²
(x - 3)² + y² - 8y + 16 = y² - 2y + 1
(x - 3)² + 15 = 6y
Divide each side by 6 to obtain
y = (1/6) (x - 3)² + 5/2
The vertex of the parabola is at (3, 5/2).
The line of symmetry is x = 3.
The leading coefficient, 1/6, is positive, therefore the curve opens upward.
Answer:
At an arbitrary point P (x,y) on the parabola, the distance, d, from the focus to P should be equal to the distance from P to the directrix.
That is,
d² = (x - 3)² + (y - 4)² = (y - 1)²
(x - 3)² + y² - 8y + 16 = y² - 2y + 1
(x - 3)² + 15 = 6y
Divide each side by 6 to obtain
y = (1/6) (x - 3)² + 5/2
The vertex of the parabola is at (3, 5/2).
The line of symmetry is x = 3.
The leading coefficient, 1/6, is positive, therefore the curve opens upward.
Answer:
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