What is the equation of the parabola with the vertex #(-2,5)# and focus #(-2,6)#?

1 Answer
Apr 2, 2017

Equation of parabola is #4y=x^2+4x+24#

Explanation:

As the vertex #(-2,5)# and focus #(-2,6)# share same abscissa i.e. #-2#, parabola has axis of symmetry as #x=-2# or #x+2=0#

Hence, equation of parabola is of the type #(y-k)=a(x-h)^2#, where #(h,k)# is vertex. Its focus then is #(h,k+1/(4a))#

As vertex is given to be #(-2,5)#, the equation of parabola is

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

  • as vertex is #(-2,5)# and parabola passes through vertex.

and its focus is #(-2,5+1/(4a))#

Therefore #5+1/(4a)=6# or #1/(4a)=1# i.e. #a=1/4#

and equation of parabola is #y-5=1/4(x+2)^2#

or #4y-20=(x+2)^2=x^2+4x+4#

or #4y=x^2+4x+24#

graph{4y=x^2+4x+24 [-11.91, 8.09, -0.56, 9.44]}