Question

The equation of a parabola is x = -4(y - 1)^2. What is the equation of the directrix?

Answer 1

http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections.faq.question.44360.html

Answer 2

keep in mind that \(\bf (y-\color{red}{k})^2=4\color{blue}{p}(x-\color{red}{h})\\ \quad \\
\textit{vertex is at }\quad (h,k)\\
\textit{directrix is "p" distance from the vertex}\\\\ \quad \\ \quad \\
x = -4(y - 1)^2\implies (x-\color{red}{0})=-4(\color{blue}{1})(y-\color{red}{1})^2\)

Answer 3

@jdoe0001 so x = -1/4

Answer 4

hmmm well, notice that the squared variable is the "y", not the "x", that means that the parabola is opening horizontally
also notice that the coefficient in front of the squared binomial, is negative, that is \(\bf x = \color{red}{-4}(y - 1)^2\) that means that the parabola opening to the left-hand-side

the directrix will be "p" distance from the vertex in the opposite direction the parabola is going....

so what's your vertex at?

Answer 5

so the parabola looks sorta like

Answer 6

16 ? @jdoe0001

Answer 7

-1/16 @jdoe0001

Answer 8

hmm well, where do you think is the coordinates for the vertex?

Answer 9

1/16

Answer 10

yeap x= 1/16 :)

Answer 11

\(\bf x = -4(y - 1)^2\implies (x-0)=4(-1)(y-1)^2\\ \quad \\\implies -\cfrac{1}{4}(x-0)=(y-1)^2\\ \quad \\
-4\cfrac{1}{16}(x-0)=(y-1)^2\)

Answer 12

I have another if you feel like helpin out lol :)