How do you find the coordinates of the two points on the #x^2-2x+4y^2+16y+1=0# closed curve where the line tangent to the curve is vertical?

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Answer 1 · 2015-02-28T18:02:33

The answer are: #P(-3,-2)# and #Q(5,-2)#.

#x^2-2x+4(y^2+4y)+1=0rArr#

#x^2-2x+1-1+4(y^2+4y+4-4)+1=0rArr#

#(x-1)^2-1+4(y+2)^2-16+1=0rArr#

#(x-1)^2+4(y+2)^2=16rArr#

#(x-1)^2/16+4(y+2)^2/16=1#

#(x-1)^2/16+(y+2)^2/4=1#,

That is the standard equation of an ellipse.

Its center is #C(1,-2)#,

and its semi-axes are:

#a=4#
#b=2#.

graph{x^2-2x+4y^2+16y+1=0 [-10, 10, -5, 5]}

The points in which the tangent is vertical are easy to find, also looking the graph:

#P(-3,-2)#

and

#Q(5,-2)#.