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Find the equation of the parabola with focus (-3,2) and directrix x-y + 1 = 0.? ​

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User Lyra
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Explanation:

To find the equation of the parabola with focus (-3,2) and directrix x-y+1=0, we can use the definition of a parabola: the set of all points that are equidistant to the focus and directrix.

Let P(x, y) be an arbitrary point on the parabola, and let d(P, directrix) be the distance from P to the directrix. The distance from P to the focus is given by the distance formula:

d(P, focus) = √[(x - (-3))^2 + (y - 2)^2] = √[(x + 3)^2 + (y - 2)^2]

Since P is equidistant from the focus and directrix, we have:

d(P, directrix) = |x - y + 1| / √(1^2 + (-1)^2) = |x - y + 1| / √2

Therefore, the equation of the parabola is given by:

d(P, focus) = d(P, directrix)

√[(x + 3)^2 + (y - 2)^2] = |x - y + 1| / √2

Squaring both sides and simplifying, we get:

(x + 3)^2 + (y - 2)^2 = (x - y + 1)^2 / 2

Expanding the right-hand side and simplifying, we get:

2(x + 3)^2 + 2(y - 2)^2 = (x - y + 1)^2

Expanding the right-hand side again and simplifying, we get:

2x^2 + 8xy + 2y^2 - 8x - 12y + 20 = 0

Therefore, the equation of the parabola is:

2x^2 + 8xy + 2y^2 - 8x - 12y + 20 = 0

which is in general form.

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User Ribs
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