Find the equation of the parabola which passe | vedclass

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(D) The vertex of the parabola is at the origin $(0,0)$ and its axis is along the $y$-axis. Thus,the equation of the parabola is of the form $x^2 = 4a

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$x^2=-18y$

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(D) The vertex of the parabola is at the origin $(0,0)$ and its axis is along the $y$-axis. Thus,the equation of the parabola is of the form $x^2 = 4ay$ or $x^2 = -4ay$. Since the parabola passes through $(6,-2)$,which lies in the fourth quadrant,the parabola must open downwards. Therefore,the equation is $x^2 = -4ay$. Substituting the point $(6,-2)$ into the equation: $(6)^2 = -4a(-2)$ $36 = 8a$ $a = \frac{36}{8} = \frac{9}{2}$. Substituting $a = \frac{9}{2}$ back into the equation $x^2 = -4ay$: $x^2 = -4 \left(\frac{9}{2}\right)y$ $x^2 = -18y$.

Find the equation of the parabola which passes through $(6,-2)$,has its vertex at the origin and its axis along the $y$-axis.

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