How do I graph the ellipse with the equation #(x−5)^2/9+(y+1)^2/16=1#?

Answer 1

From the equation, by inspection, Center#(5, -1)# length of Semi-major axis#=4#, length of Semi-minor axis#=3#, Vertices at #(5, 3)# and #(5, -5)#, Co-vertices at #(8, -1)# and #(2, -1)#

This is an Ellipse with Vertical Major Axis

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

so that #a^2=16# and #a=4# and #b^2=9# and #b=3#

so we can calculate, Vertices using the values of #a, b, h, k#

Vertex #V_1(h, k+a)=V_1(5, -1+4)=V_1(5, 3)# Vertex #V_2(h, k-a)=V_2(5, -1-4)=V_2(5, -5)#

Co-vertex #V_3(h+b, k)=V_3(5+3,-1)=V_3(8, -1)# Co-vertex #V_4(h-b, k)=V_3(5-3, -1)=V_3(2, -1)#

We don't use the Foci to graph the ellipse so there's no need to compute for them

Kindly see the graph of #(x-5)^2/9+(y+1)^2/16=1# and locate the vertices and co-vertices

graph{(x-5)^2/9+(y+1)^2/16=1[-12,12,-6,6]}

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.