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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 (5,2) and symmetric with respect to y-axis

$\begin {array} {1 1} (A)\;2x^2=25y & \quad (B)\;2x^2=-25y \\ (C)\;2y^2=25x & \quad (D)\;2y^2=-25x \end {array}$

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  • Equation of the parabola along y - axis and opening upwards is $x^2=4ay$
  • Equation of the parabola along y - axis and opening downwards is $ x^2=-4ay$
The given vertex is (0,0) and
the parabola is symmetric about the y - axis.
Hence the equation of the parabola is either of the form.
$x^2=4ay $ or $ x^2 = -4ay$
The curve passes through the point (5,2) .
Now substituting for x and y we get,
$(5)^2=-4a(2)$
$ \Rightarrow 25 = -8a$ or
$a = -\large\frac{25}{8}$
Hence the equation of the parabola is
$ x^2=-4 \bigg( \large\frac{25}{8} \bigg)$$y$
(i.e) $x^2=-\large\frac{25}{2}$$y$
or $2x^2=-25y$ is the required equation of the parabola.
answered Jul 15, 2014 by thanvigandhi_1
 

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