How do you find the foci, center, vertex for #(1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1#?

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Answer 1 · 2016-01-27T10:30:06

#Foci: F_1(-2+sqrt7,5), F_2(-2-sqrt7, 5)# and #Center:(-2, 5)#
#Vertex: V_1(2, 5), V_2(-6, 5)#

From the given equation of ellipse:

#(x-h)^2/a^2+(y-k)^2/b^2=1#

From #(x+2)^2/16+(y-5)^2/9=1#

it follows that

#(x--2)^2/4^2+(y-5)^2/3^2=1#

And the #Center (h, k)=(-2, 5)#

Compute distance from center to focus #c#

#c=sqrt(a^2-b^2)=sqrt(16-9)=sqrt7#

Compute #Foci# # F_1, F_2#

#F_1(-2+sqrt7, 5)#

#F_2(-2-sqrt7, 5)#

Compute #Vertices# # V_1, V_2#

#V_1(-2+a, 5)=V_1(-2+4, 5)=V_1(2, 5)#

#V_2(-2-a, 5)=V_2(-2-4, 5)=V_2(-6, 5)#

Have a nice day !! from the Philippines!