Question

Find the equation of the parabola with vertex at the origin, passing through the point $P(3,-4)$ and symmetric about the y-axis.

Answer 1

It is given that the vertex of the parabola is $O(0,0)$ and it is symmetric about the y-axis.

So, its equation is $x^2=4ayor{x}^2=-4ay.$

Since the parabola passes through the point $P(3,-4),$ so it lies in the 4th quadrant.

$\therefore$ it is a downward parabola.

Let its equation be $x^2=-4ay.$

Since it passes through the point $P(3,-4),$

we have

$3^2=-4\times a\times (-4)\Rightarrow a=\frac{9}{16}.$

So, the required equation is

$x^2=-4\times \frac{9}{16}y\Rightarrow x^2=\frac{-9}{4}y\Rightarrow 4x^2+9y=0.$