What are the vertex, focus and directrix of # y=x^2+10x+21 #?

1 Answer
Mar 4, 2016

Vertex is #-5,-4)#, (focus is #(-5,-15/4)# and directrix is #4y+21=0#

Explanation:

Vertex form of equation is #y=a(x-h)^2+k# where #(h,k)# is vertex

The given equation is #y=x^2+10x+21#. It may be noted that the coefficient of #y# is #1# and that of #x# too is #1#. Hence, for converting the same, we have to make terms containing #x# a complete square i.e.

#y=x^2+10x+25-25+21# or

#y=(x+5)^2-4# or

#y=(x-(-5))^2-4#

Hence vertex is #(-5,-4)#

Standard form of parabola is #(x - h)^2=4p(y - k)#,

where focus is #(h,k+p)# and directrix #y=k-p#

As the given equation can be written as #(x-(-5))^2=4xx1/4(y-(-4))#, we have vertex #(h,k)# as #(-5,-4)# and

focus is #(-5,-15/4)# and directrix is #y=-5-1/4=-21/4# or #4y+21=0#