CALC
posted by Adrianna .
point on the hyperbola 9x^2−6y^2=10 closest to the point (0, 7).
y coordinate of each point= ?
positive x coordinate= ?
negative x coordinate= ?

let the point be P(a,b)
slope of line form P to (0,7) = (b7)/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 (b7)/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