find equations for the tangent

Question

and the normal at P (1n2, 2k)
on the curve with equation
y=ke^x, where k is a constant.
(b) given that y=sin^-1x, show
that (1-x^2)d^2y/dx^2-xdy/ dx=0

Answer 1

(a) no idea what 1n2 means

(b)
y = arcsin(x)
y' = 1/√(1-x^2)
y" = x/(1-x^2)^3/2

so,

(1-x^2)y" - xy'
= x/√(1-x^2) - x/√(1-x^2)
= 0

Answer 2

(a) Ah. I think that's ln2 (not 1n2)
y' = ke^x
y'(ln2) = 2k
So, now we have a point and a slope. Back to algebra I:

tangent: y-2k = 2k(x-ln2)
normal: y-2k = -1/2k (x-ln2)