We will discuss about the standard form of parabola x\(^{2}\) = -4ay
Equation y\(^{2}\) = -4ax (a > 0) represents the equation of a parabola whose co-ordinate of the vertex is at (0, 0), the co-ordinates of the focus are (0, -a), the equation of directrix is y = a or y - a = 0, the equation of the axis is x = 0, the axis is along negative y-axis, the length of its latus rectum = 4a and the distance between its vertex and focus is a.
Solved examples based on the standard form of parabola x\(^{2}\) = -4ay:
1. Find the axis, co-ordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola x\(^{2}\) = -16y
Solution:
The given parabola x\(^{2}\) = -16y
⇒ x\(^{2}\) = -4 ∙ 4 y
Compare the above equation with standard form of parabola x\(^{2}\) = -4ay, we get, a = 4.
Therefore, the axis of the given parabola is along negative y-axis and its equation is x = 0
The co-ordinates of its vertex are (0, 0) and the co-ordinates of its focus are (0, -4); the length of its latus rectum = 4a = 4 ∙ 4 = 16 units and the equation of its directrix is y = a i.e., y = 4 i.e., y - 4 = 0.
2. Find the axis, co-ordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola 3x\(^{2}\) = -8y
Solution:
The given parabola 3x\(^{2}\) = -8y
⇒ x\(^{2}\) = -\(\frac{8}{3}\)y
⇒ x\(^{2}\) = -4 ∙ \(\frac{2}{3}\) y
Compare the above equation with standard form of parabola x\(^{2}\) = -4ay, we get, a = \(\frac{2}{3}\).
Therefore, the axis of the given parabola is along negative y-axis and its equation is x = 0
The co-ordinates of its vertex are (0, 0) and the co-ordinates of its focus are (0, -\(\frac{2}{3}\)); the length of its latus rectum = 4a = 4 ∙ \(\frac{2}{3}\) = \(\frac{8}{3}\) units and the equation of its directrix is y = \(\frac{2}{3}\) i.e., 3y = 2 i.e., 3y - 2 = 0.
● The Parabola
11 and 12 Grade Math
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