Question 145226
What is the directrix of the parabola with the equation y+3= 1/10 (x+2)^2?
<pre><font size = 4 color = "indigo"><b>
What you have to know about parabolas in standard form:

1: Parabolas whose equations are in the standard form

{{{(x-h)^2 = 4p(y-k)}}}

opens upward if {{{p > 0}}} and downward if {{{p < 0}}}

They have vertex (h, k), focus (h, k+p), and
the directrix is the horizontal line whose equation
is y=k-p

2: Parabolas whose equations are in the standard form

{{{(y-k)^2 = 4p(x-h)}}}

opens rightward if {{{p > 0}}} and leftward if {{{p < 0}}}

They have vertex (h, k), focus (h+p, k), and
the directrix is the vertical line whose equation
is x=h-p.


Your equation

{{{y+3= (1/10) (x+2)^2}}}

can be placed in the standard form 1.

Multiply both sides by 10

{{{10(y+3)= 10*(1/10)(x+2)^2}}}

{{{10(y+3)= cross(10)*(1/cross(10))(x+2)^2}}}

{{{10(y+3)= (x+2)^2}}}

Swap sides:

{{{(x+2)^2=10(y+3)}}}

Compare to the standard equation in 1 above:

{{{(x-h)^2 = 4p(y-k)}}}

So {{{-h=2}}} so {{{h=-2}}},
{{{4p=10}}} so {{{p=10/4=5/2}}}
{{{-k=3}}} so {{{k=-3}}}
opens upward because {{{p = 5/2 > 0}}} 

It has vertex (h, k) = (-2,-3)

It has focus (h, k+p) = (-2,-3+{{{5/2}}}) = (-2, {{{-1/2}}})
the directrix is the horizontal line whose equation
is {{{y=k-p}}} or {{{y = -3-(5/2)}}} or {{{y = -11/2}}}

To draw the graph, plot the focus (-2, {{{-1/2}}}), the vertex(-2,-3) and
the directrix {{{y=-11/2}}}

{{{drawing(500,500,-10,6,-8,8,
locate(-2-.1,-.5+.25,o),locate(-2-.1,-3+.25,o),
graph(500,500,-10,6,-8,8,0,-11/2)

 )}}} 

Draw a line from the focus directly to the directrix. The
vertex should be the midpoint of this line.

{{{drawing(500,500,-10,6,-8,8,
locate(-2-.1,-.5+.25,o),locate(-2-.1,-3+.25,o),line(-2,-1/2,-2,-11/2),
graph(500,500,-10,6,-8,8,0,-11/2)

 )}}}

Draw a square with that line being its left side:

{{{drawing(500,500,-10,6,-8,8,
locate(-2-.1,-.5+.25,o),locate(-2-.1,-3+.25,o),line(-2,-1/2,-2,-11/2),
graph(500,500,-10,6,-8,8,0,-11/2),
rectangle(-2,-11/2,-2+5,-11/2+5)


 )}}}

Draw another square with that line being its right side:

{{{drawing(500,500,-10,6,-8,8,
locate(-2-.1,-.5+.25,o),locate(-2-.1,-3+.25,o),line(-2,-1/2,-2,-11/2),
graph(500,500,-10,6,-8,8,0,-11/2),
rectangle(-2,-11/2,-2+5,-11/2+5),
rectangle(-2,-11/2,-2-5,-11/2+5)

 )}}}

Finally, draw the parabola through the vertex and the
upper corners of the two squares:

{{{drawing(500,500,-10,6,-8,8,
locate(-2-.1,-.5+.25,o),locate(-2-.1,-3+.25,o),line(-2,-1/2,-2,-11/2),
graph(500,500,-10,6,-8,8,.1(x+2)^2-3,-11/2),
rectangle(-2,-11/2,-2+5,-11/2+5),
rectangle(-2,-11/2,-2-5,-11/2+5)

 )}}}

Edwin</pre>