Find the equation of the parabola whose axis is parallel to the $X$ axis and the parabola passes through $\left(3, 3\right)$, $\left(6, 5\right)$ and $\left(6, -3\right)$.
The axis of the required parabola is parallel to the $X$ axis.
$\therefore \;$ Let the equation of the parabola be
$x = Ay^2 + By + C$ $\;\;\; \cdots \; (1)$
Since equation $(1)$ passes through the points $\left(3, 3\right)$, $\left(6, 5\right)$ and $\left(6, -3\right)$, we have
$3 = 9A + 3B + C$ $\;\;\; \cdots \; (2a)$
$6 = 25 A + 5B + C$ $\;\;\; \cdots \; (2b)$
$6 = 9A - 3B + C$ $\;\;\; \cdots \; (2c)$
Solving equations $(2a)$ and $(2c)$ simultaneously gives
$- 3 = 6 B$ $\implies$ $B = \dfrac{-1}{2}$
Solving equations $(2a)$ and $(2b)$ simultaneously gives
$3 = 16A + 2B$ $\;\;\; \cdots \; (3a)$
Substituting the value of $B$ in equation $(3a)$ gives
$3 = 16A + 2 \times \left(\dfrac{-1}{2}\right)$
i.e. $\;$ $4 = 16 A$ $\implies$ $A = \dfrac{1}{4}$
Substituting the values of $A$ and $B$ in equation $(2a)$ gives
$3 = 9 \times \dfrac{1}{4} + 3 \times \left(\dfrac{-1}{2}\right) + C$
i.e. $\;$ $C = 3 - \dfrac{9}{4} + \dfrac{3}{2} = \dfrac{9}{4}$
Substituting the values of $A$, $B$ and $C$ in equation $(1)$ gives
$x = \dfrac{1}{4} y^2 - \dfrac{1}{2} y + \dfrac{9}{4}$
i.e. $\;$ $y^2 - 2y - 4x + 9 = 0$ $\;\;\; \cdots \; (4)$
Equation $(4)$ is the required equation of the parabola.