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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=6y$

$\begin {array} {1 1} (A)\;focus : \bigg(0, -\large\frac{3}{2} \bigg), Equation \: of \: directrix : 2y-3=0 , Axis : y - axis , Length \: of \: the \: L.R : 6 \\ (B)\;focus : \bigg( -\large\frac{3}{2}, 0 \bigg), Equation \: of \: directrix : 2x+3=0 , Axis : x - axis , Length \: of \: the \: L.R : 6 \\ (C)\; focus : \bigg(0, \large\frac{3}{2} \bigg), Equation \: of \: directrix : 2y+3=0 , Axis : y - axis , Length \: of \: the \: L.R : 6\\ (D)\;focus : \bigg( \large\frac{3}{2}, 0 \bigg), Equation \: of \: directrix : 2x-3=0 , Axis : x - axis , Length \: of \: the \: L.R : 6 \end {array}$

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Toolbox:
  • Equation of the parabola $x^2=4ay$
  • If the coefficient of y is positive, the parabola opens upwards.
  • If the equation involves $x^2$, then the axis of the parabola is y - axis.
  • Equation of the directrix is y = -a or y+a=0.
  • Length of the latus rectum is 4a.
Step 1 :
The given equation is $x^2=6y$
Since the coefficient of y is positive. Hence the parabola opens upwards.
Comparing this equation with $x^2 = 4ay$
$4a=6$
$ \Rightarrow a = \large\frac{6}{4}$$ = \large\frac{3}{2}$
Hence the coordinates of the focus = (0, a) = $ \bigg( 0, \large\frac{3}{2} \bigg)$
Equation of the directrix $y= - \large\frac{3}{2}$ or $y+\large\frac{3}{2}=0$
(i.e) $2y+3=0$
Length of the latus rectum $4a= 4 \times \large\frac{3}{2}$$=6$
Axis : y - axis.
answered Jul 14, 2014 by thanvigandhi_1
 

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