What is the equation in standard form of the parabola with a focus at (14,5) and a directrix of y= -15?

2 Answers
Dec 30, 2017

The equation of parabola is #y=1/40(x-14)^2-5 #

Explanation:

Focus is at #(14,5) #and directrix is #y=-15#. Vertex is at midway

between focus and directrix. Therefore vertex is at

#(14,(5-15)/2) or (14, -5)# . The vertex form of equation of

parabola is #y=a(x-h)^2+k ; (h.k) ;# being vertex. Here

# h=14 and k = -5 # So the equation of parabola is

#y=a(x-14)^2-5 #. Distance of vertex from directrix is

#d= 15-5=10#, we know # d = 1/(4|a|):. |a|=1/(4d)# or

#|a|=1/(4*10)=1/40# . Here the directrix is below

the vertex , so parabola opens upward and #a# is positive.

#:. a=1/40# Hence the equation of parabola is

#y=1/40(x-14)^2-5 #
graph{1/40(x-14)^2-5 [-90, 90, -45, 45]} [Ans]

Dec 30, 2017

#(x-14)^2=40(y+5)#

Explanation:

#"the standard form of a parabola in "color(blue)"translated form"# is.

#•color(white)(x)(x-h)^2=4p(y-k)#

#"where "(h,k)" are the coordinates of the vertex"#

#"and p is the distance from the vertex to the focus"#

#"since the directrix is below the focus then the curve"#
#"opens upwards"#

#"coordinates of vertex "=(14,(5-15)/2)=(14,-5)#

#"and "p=5-(-5)=10#

#rArrrArr(x-14)^2=40(y+5)larrcolor(red)"equation of parabola"#