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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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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 for the following : $ x^2=-16y$

$\begin {array} {1 1} (A)\;focus (0, 4) , Axis : y - axis, Equation \: of \: directrix : y+4=0, Length \: of \: the \: latus \: rectum : 16 \\ (B)\; focus (0, -4) , Axis : y - axis, Equation \: of \: directrix : y-4=0, Length \: of \: the \: latus \: rectum : 16 \\ (C)\;focus (4, 0) , Axis : x - axis, Equation \: of \: directrix : x+4=0, Length \: of \: the \: latus \: rectum : 16 \\ (D)\;focus (-4, 0) , Axis : x - axis, Equation \: of \: directrix : x-4=0, Length \: of \: the \: latus \: rectum : 16 \end {array}$

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  • Equation of the parabola $x^2=-4ay$
  • Since the coefficient of y is negative the curve is open downwards.
  • Coordinates of focus is (0, -a)
  • Length of the latus rectum is 4a.
  • Equation of directrix is y-a = 0
Step 1 :
Given equation of the parabola is $x^2=-16y$
Since the coefficient of y is negative the curve is open downwards.
Comparing this equation with $x^2=-4ay$
$4a=16 \Rightarrow a = 4$
Hence the coordinates of focus is (0, -4)
The given equation involves $x^2$,
Axis is y - axis.
Equation of directrix is y = 4 or y-4=0
Length of the latus rectum is $4 \times 4 = 16$
answered Jul 14, 2014 by thanvigandhi_1
 

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