Question

What is the directrix of the parabola with the equation y+3=1/10(x+2)squared?

Answer 1

site is a bit laggy, thus

\(\bf 4p(y-{\color{blue}{ k}})=(x-{\color{brown}{ h}})^2\qquad vertex\ ( {\color{brown}{ h}},{\color{blue}{ k}})\\ \quad \\ \quad \\

y+3=\cfrac{1}{10}(x+2)^2\implies y+3=\cfrac{(x+2)^2}{10}\implies 10(y+3)=(x+2)^2\)
see the vertex of the parabola?

Answer 2

No I'm sorry.

Answer 3

\(\bf y+3=\cfrac{1}{10}(x+2)^2\implies y+3=\cfrac{(x+2)^2}{10}\implies 10(y+3)=(x+2)^2
\\ \quad \\
10(y-(-3))=(x-(-2))^2\)

what about now?