What is the equation for a parabola with a vertex at (5,-1) and a focus at (5,-7)?

2 Answers
Jan 13, 2018

#y=-1/24x^2+5/12x-49/24#

Explanation:

#"note that the vertex and focus are on the vertical axis"#

#x=5#

#"the focus and directrix are equidistant from the vertex"#

#"that is "-1-(-7)=6#

#"equation of directrix is "y=-1+6toy=5#

#"for any point "(x,y)" on the parabola"#

#"the distance to the focus and directrix are equidistant"#

#"using the "color(blue)"distance formula"#

#sqrt((x-5)^2+(y+7)^2)=|y-5|#

#color(blue)"squaring both sides"#

#(x-5)^2+(y+7)^2=(y-5)^2#

#(y+7)^2-(y-5)^2=-(x-5)^2#

#cancel(y^2)+14y+49cancel(-y^2)+10y-25=-x^2+10x-25#

#rArr24y=-x^2+10x-49#

#rArry=-1/24x^2+5/12x-49/24#

Jan 13, 2018

#-24(y+1)=(x-5)^2#

Explanation:

Let's graph the points we have.

Desmos

The vertex and focus have the same x-coordinate. This tells us that this parabola doesn't open sideways, which means we'll be using the equation #4p(y-k)=(x-h)^2#
(Here's a link if you want to see how this equation was derived!)

In the equation, #p# is the distance from the vertex to the focus or directrix; #k# is the y-coordinate of the vertex, and #h# is the x-coordinate of the vertex.

Zhirou

Since the vertex is already given, we can plug that in:
#4p(y-k)=(x-h)^2#
#4p(y+1)=(x-5)^2#

To find the p-value, simply subtract the y-value of the vertex from the y-value of the focus.
#-7+1 = -6#
The p-value is negative. This makes sense, because the focus of the parabola is below the vertex, making the parabola open downwards. Downward parabolas have negative p-values.

Now, just plug that into the equation:
#4(-6)(y+1)=(x-5)^2#
#-24(y+1)=(x-5)^2#
Desmos