Question 970161
Find and equation for the hyperbola that satisfies the given conditions.
Vertices: (0,14), (0,-14) and passes through (-5,21)
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Given hyperbola has a vertical transverse axis and center at the origin.
Its standard form of equation: {{{y^2/a^2-x^2/b^2=1}}}
a=14 (distance from center to vertices)
a^2=196
{{{y^2/a^2-x^2/b^2=1}}}
solve for b^2
plug-in coordinates of given point on the curve(-5, 21)
{{{21^2/14^2-(-5^2)/b^2=1}}}
{{{21^2/196-25/b^2=1}}}
{{{2.25-25/b^2=1}}}
25/b^2=2.25-1=1.25
b^2=25/1.25
b^2=20
equation: {{{y^2/196-x^2/20=1}}}