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Q.
Find the equation of the parabola which is symmetric about the $y$-axis, and passes through the point $(2,-3)$.
Conic Sections
Solution:
Since the parabola is symmetric about $y$-axis and has its vertex at the origin so the equation is of the form $x^{2} = 4ay$ or $x^{2} = - 4ay$, where the sign depends on whether the parabola opens upwards or downwards. But the parabola passes through $(2, -3)$ which lies in the fourth quadrant so it must open downwards. Thus the equation is of the form $x^{2} = - 4ay$.
Since the parabola passes through $(2, -3)$, we have
$2^{2} = - 4a(-3)$,
i.e., $a=\frac{1}{3}$
Therefore, the equation of the parabola is
$x^{2}=-4\left(\frac{1}{3}\right)y$,
i.e., $3x^{2}=-4y$.