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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

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

Toolbox:
• 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.