Asked by Marcos
find equation of tangent common to x^2-(y^2)/3=1 and y^2=8x.
Answers
Answered by
Steve
hmmm. we want the line passing through (a,b) and (c,d) such that
c^2 - d^2/3 = 1
b^2 = 8a
(b-d)/(a-c) = 4/b
(b-d)/(a-c) = 3c/d
Due to symmetry, there will be two tangent lines, but one solution is
(a,b) = (1/2,2)
(c,d) = (-2,-3)
See the graphs at
http://www.wolframalpha.com/input/?i=plot+x^2-%28y^2%29%2F3%3D1+%2C+y^2%3D8x%2C+y+%3D+2%28x-1%2F2%29%2B2
c^2 - d^2/3 = 1
b^2 = 8a
(b-d)/(a-c) = 4/b
(b-d)/(a-c) = 3c/d
Due to symmetry, there will be two tangent lines, but one solution is
(a,b) = (1/2,2)
(c,d) = (-2,-3)
See the graphs at
http://www.wolframalpha.com/input/?i=plot+x^2-%28y^2%29%2F3%3D1+%2C+y^2%3D8x%2C+y+%3D+2%28x-1%2F2%29%2B2
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