Posted by Adrianna on Wednesday, April 6, 2011 at 10:48pm.
let the point be P(a,b)
slope of line form P to (0,7) = (b-7)/a
differentiate:
18x - 12y dy/dx = 0
dy/dx = 18x/(12y) = 3x/(2y)
at P, dy/dx = 3a/(2b)
so the slope of the tangent at P is 3a/(2b)
by basic geometry the slope of that tangent and the slope of the line to (0,7) to P must be negative reciprocals of each other.
so (b-7)/a = -2b/(3a)
-2ab = 3ab - 21a
5ab = 21a
b = 21/5
sub into 9a^2 - 6b^2 = 10
9a^2 - 6(441/25) = 10
a = ± 4√181/15
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