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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of a parabola that satisfies the following condition - vertex $(0,0)$ and passing through $(5,2)$ and symmetric with respect to y-axis.

$\begin{array}{1 1} x^2= \frac{25}{2} y \\y^2= \frac{25}{8} x \\x^2= \frac{-25}{4} y \\ y^2= \frac{-25}{16} x \end{array} $

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Given vertex is origin and axis is y-axis, the equation is of the form $x^2 = 4ay $ of $x^2 = -4ay$.
Since the parabola passes through $(5,2)$, the equation is of the form $x^2=4ay$ which means that $5^2 = 4 \times a \times 2 \rightarrow a = \large\frac{25}{8}$
Therefore, the equation of the parabola is $x^2 =4 \large\frac{25}{8}$$y = \large\frac{25}{2}$$ y$
answered Apr 1, 2014 by balaji.thirumalai
 

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