Recognition of conics:
The general equation of 2nd degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents
A pair of straight line If Δ= 0 where D = abc + 2fgh - af2 - bg2 - ch2 , h2 ≥ ab
A circle if Δ ≠ 0, h = 0, a = b
A parabola if Δ ≠ 0, ab - h2 = 0
An ellipse if Δ ≠ 0, ab - h2 > 0
An hyperbola if Δ ≠ 0, ab - h2 < 0
Example: Find equation of the parabola whose focus is (3, -4) and directrix is x - y + 5 = 0.
Solution: Let P(x, y) be any point on parabola. Then
=> x2 + y2 + 2xy - 22x + 26y + 25 = 0 => (x + y)2 = 22x - 26y - 25.
Example: If (0, 4) and (0, 2) are respectively vertex and focus of a parabola, then its equation is
(A) x2 + 8y = 32 (B) y2 + 8x = 32
(C) x2 - 8y = 32 (D) y2 - 8x = 32
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Solution: AS = 2 = a. Vertex (0, 4) lies on y-axis. Thus the parabola X2 = -4aY is a downward parabola as focus is below vertex.
Or (x -0)2 = -4 x 2(y -4)
Or x2 + 8y = 32
Hence (A) is the required answer.
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