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Find the co-ordinates of the focus, axis of the parabola, equations of the directrix and length of latus rectum of the parabola $x^2=6y$

\begin{array}{1 1}(0,\large\frac{3}{2}) \quad \text{y-axis} \quad y = \large\frac{-3}{2} \quad 6 \\ (0,\large\frac{3}{2}) \quad \text{y-axis} \quad y = \large\frac{-1}{2} \quad 6 \\ (0,\large\frac{-3}{2}) \quad \text{x-axis} \quad y = \large\frac{-3}{2} \quad 6 \\ (0,\large\frac{-3}{2}) \quad \text{x-axis} \quad y = \large\frac{-3}{2} \quad 6\end{array}

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Given $x^2=6y$
The co-efficient of $y$ is positive, comparing with $x^2 = 4ay \rightarrow 4a = 6 \rightarrow a = \large\frac{3}{2}$
1) Therefore co-ordinates of the focus = $(0,a) = (0,\large\frac{3}{2})$
2) Since $x^2 = 6x$, the axis of the parabola is the y-axis
3) The Equation of the directrix $ y = -a \rightarrow y = \large\frac{-3}{2}$
4) The Length of the Latus Rectum $ = 4a = 4 \times \large\frac{3}{2}$ = 6


answered Apr 1, 2014 by balaji.thirumalai
edited Apr 1, 2014 by balaji.thirumalai

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