How do you graph #y^2/100-x^2/75=1# and identify the foci and asympototes?

1 Answer
Nov 11, 2016

Answer:

The foci are F#=(0,5sqrt7)# and F'#=(0,-5sqrt7)#
The asymptotes are #y=(2x)/sqrt3# and #y=-(2x)/sqrt3#

Explanation:

This is an up-down hyperbola.
The general equation is #y^2/a^2-x^2/b^2=1#
The center is #(0,0)#
To determine the foci, we calculate #c=sqrt(a^2+b^2)=+-sqrt175=+-5sqrt7#
The foci are F#=(0,+5sqrt7)# and F'#=(0,-5sqrt7)#

The vertices are #(0,10)# and #(0,-10)#

The slopes of the asymptotes are #+-10/sqrt75=+-10/(5sqrt3)=+-2/sqrt3#
The equations of the asymptotes are #y=(2x)/sqrt3# and #-(2x)/sqrt3#
graph{((y^2/100)-(x^2/75)-1)(y-(2x/sqrt3))(y+(2x/sqrt3))=0 [-74, 74.1, -37.07, 37.06]}