Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
$x^{2} = - 16 y$

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Solution

The given equation of parabola is x^2 = - 16y
which is of the form $x^{2} = - 4 a y ∴ 4 a = 6 \Rightarrow a = 4$
$∴$ Coordinates of foucs are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is

$y = 4 \Rightarrow y - 4 = 0$
Length of latus rectum $= 4 \times 4 = 16.$

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Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x² = – 16y

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Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. x2=−16y

The given equation of parabola is x^2 = - 16y which is of the form x2=−4ay∴4a=6⇒a=4 ∴ Coordinates of foucs are (0, -4) Axis of parabola is x = 0 Equation of the directrix is y=4⇒y−4=0 Length of latus rectum =4×4=16.

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
$x^{2} = - 16 y$

The given equation of parabola is x^2 = - 16y
which is of the form $x^{2} = - 4 a y ∴ 4 a = 6 \Rightarrow a = 4$
$∴$ Coordinates of foucs are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is

$y = 4 \Rightarrow y - 4 = 0$
Length of latus rectum $= 4 \times 4 = 16.$