Chapter 4 - Review Exercises - Page 353: 50a

$(0, 45)$

$y=ax^2$ $20=a(-60)^2$ or, $a =\dfrac{20}{3600}=\dfrac{1}{180}$ So, $y=\dfrac{x^2}{180}$ We will write the equation in standard form as follows: $ x^2= (4)(45)(y)$ We can see that $p=45$. So, the coordinates of the focus are: $(0, 45)$