Asked by Babji
At what point(s) does 4x^2+y^2−8x+4y+4=0 have horizontal and/or vertical tangent line(s)?
A. (1,0),(1,−4),(0,−2),(2,−2)
B. (1,−2),(2,−1)
C. (1,0),(0,−2)
D. (1,−2)
A. (1,0),(1,−4),(0,−2),(2,−2)
B. (1,−2),(2,−1)
C. (1,0),(0,−2)
D. (1,−2)
Answers
Answered by
oobleck
using implicit derivatives, 8x + 2yy' - 8 + 4y' = 0
y' = -4(x-1)/(y+2)
so, horizontal tangents where x-1=0
vertical tangents where y+2=0
Note that this is just the ellipse
(x-1)^2/4 + (y+2)^2/1 = 1
and the tangents in question are at the vertices and covertices
y' = -4(x-1)/(y+2)
so, horizontal tangents where x-1=0
vertical tangents where y+2=0
Note that this is just the ellipse
(x-1)^2/4 + (y+2)^2/1 = 1
and the tangents in question are at the vertices and covertices
Answered by
Babji
Thanks for the quick response. so would the answer be option A?
Answered by
oobleck
A is good.
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