Find the equation of the parabola which is sy | vedclass

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(A) Since the parabola is symmetric about the $y$-axis and has its vertex at the origin,the equation is of the form $x^2 = 4ay$ or $x^2 = -4ay$.Becaus

Vedclass · Accepted Answer

$3x^2 = -4y$

Vedclass · Answer

(A) Since the parabola is symmetric about the $y$-axis and has its vertex at the origin,the equation is of the form $x^2 = 4ay$ or $x^2 = -4ay$.Because the parabola passes through the point $(2, -3)$,which lies in the fourth quadrant,it must open downwards.Thus,the equation is of the form $x^2 = -4ay$.Substituting the point $(2, -3)$ into the equation:$2^2 = -4a(-3)$$4 = 12a$$a = \frac{4}{12} = \frac{1}{3}$.Substituting $a = \frac{1}{3}$ back into the equation:$x^2 = -4(\frac{1}{3})y$$x^2 = -\frac{4}{3}y$$3x^2 = -4y$.

Find the equation of the parabola which is symmetric about the $y$-axis and passes through the point $(2, -3)$.

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