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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the parabola that satisfies the given conditions : Vertex (0,0) passing through (2,3) and axis is along x-axis

$\begin {array} {1 1} (A)\;2y^2=9x & \quad (B)\;2y^2=-9x \\ (C)\;2x^2=9y & \quad (D)\;2x^2=-9y \end {array}$

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  • Equation of a parabola along the x - axis and opening rightward is $ y^2=4ax$
  • Equation of a parabola along the x - axis and opening leftward is $ y^2=-4ax$
The given vertex is (0,0)
It is also given that the axis of the parabola is x - axis.
Hence the equation of the parabola should be of the form $y^2=4ax$ or $y^2=-4ax$
The parabola passes through the point (2, 3)
Now substituing for x and y we get
$(3)^2=4a(2)$
$ \Rightarrow 9 = 8a$
$ \therefore a = \large\frac{9}{8}$
Hence the equation of the parabola is
$y^2=4 \bigg( \large\frac{9}{8} \bigg)$$x$
(i.e) $y^2=\large\frac{9}{2}$$x$
or $2y^2=9x$ is the required equation of the parabola.
answered Jul 15, 2014 by thanvigandhi_1
 

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