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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 : $x^2=6y$

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

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.