What are the vertices of the ellipse given by the equation #4x^2 + y^2 – 80x – 6y + 309 = 0#?

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Answer 1 · 2015-05-01T15:33:12

In this way, since the equation of the ellipse is this:

#(x-x_c)^2/a^2+(y-y_c)^2/b^2=1#,

#4x^2+y^2–80x–6y+309=0rArr#

#4(x^2-20x)+(y^2-6y)+309=0rArr#

#4(x^2-20x+100-100)+(y^2-6y+9-9)+309=0rArr#

#4(x^2-20x+100)-400+(y^2-6y+9)-9+309=0rArr#

#4(x-10)^2+(y-3)^2=100#

#(x-10)^2/25+(y-3)^2/100=1rArr#

So the center is in #C(10,3)# and the semi-axes are #a=5# and #b=10#.

The vertices are from the center horizontally right of #a#, horizontally left of #a#, verticalli up of #b#, vertically down of #b#.

So:

#V_1(15,3)#

#V_2(5,3)#

#V_3(10,13)#

#V_4(10,-7)#.

The graph is:

graph{(x-10)^2/25+(y-3)^2/100=1 [-15.96, 29.67, -8.02, 14.78]}