Parabolas

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Write the equation of the *parabola* $x=y^{2}-16y-10$ in the *standard form*. Then determine the *vertex*, *focus*, *directrix* and the *axis of symmetry* of the *parabola*. Also, state which way the parabola opens.

**Answer**

$x=y^{2}-16y-10$

$x=\left(y^{2}-16y\right)-10$

$x=\left(y^{2}-16y+64\right)-64-10$

$x=\left(y-8\right)^2-74$

$x+74= \left(y-8\right)^2$

The *parabola* opens to the right.

The *vertex* is given by $V\left(-74,8\right)$

The *focus* is given by $F\left(-\frac{295}{4},8\right)$

The *directrix* is given by $x= -\frac{297}{4}$

The *axis of symmetry* is given by $y=8$