4y^2-√(x)=32 (16,3) find the equation of the tangent line at the given point

Question

find the equation of the tangent line at the given point of the curve

Answer 1

4y^2-sqrt(x)=32, y^2=(32+sqrt(x))4, y=sqrt(32+sqrt(x))/2.

Let z=(32+sqrt(x))/4, then y=sqrt(z)=z^(1/2) and dy/dz=1/2sqrt(z) and dz/dx=1/(8sqrt(x)), so:

dy/dx=dy/dz*dz/dx=(1/2sqrt(z))(1/(8sqrt(x)))=1/(16sqrt(xz))=1/(8sqrt(32x+xsqrt(x)).

When x=16, y=3 and dy/dx=1/8sqrt(16(32+4))=1/8*24=1/192=tangent or slope at (16,3).

The equation of the tangent line has a slope of 1/192, so y=x/192+b, where b is y intercept. Plugging in (16,3):

3=16/192+b, b=3-1/12=35/12 and y=x/192+35/12, or 192y=x+560, the equation of the tangent line.