Determine the value of 'a' so that the curve y = (ax^2 - 4)^1/2 has a tangent with a slope of 2 at the point where x = 2.

Answer 1

y' = ax(ax²- 4)^-1/2; slope at point with x = 2 is m = 2a(4a - 4)^-1/2 = 2; a = 2√(a - 1); a² = 4a - 4; (a - 2)² = 0;

a = 2.

Answer 2

square both sides to get

y^2 =ax^2 -4

ax^2-y^2 = 4 is a hyperbola

slope of the tangent line at the point (2, sqr(4a-4)) is

y' = (1/2)(ax^2-4)^(-1/2)(2ax) = = (4a/2)/sqr(4a-4) = 2

a/sqr(4a-4) = 1

a =sqr(4a-4)

square both sides

a^2 = 4a-4

a^2-4+4 = 0

(a-2)^2 = 0

a-2 = 0

a =2

at a=2, slope = y' = (1/2)(8-4)^(-1/2)(8) = 4/2 = 2

at x=2, y= (8-4)^1/2 = 2

the point where =2 is (2,2)

y=(2x^2-4)^1/2 is the hyperbola with tangent line slope = 2 at (2,2)

or 2x^2 - y^2 = 4

or x^2/2 - y^2/4 = 1

but these equations have two branches of the hyperbola

with the fractional exponent, you only get where y>0, the top half of the two branches

vertices are (-sqr2,0), and (sqr2,0), foci are (-sqr6,0) and (sqr6,0)

asymptotes are +xsqr2 and -xsqr2

transverse is the x-axis