Find the positive value of the parameter t corresponding to a point on the curve parametrized by

Question

{x= t^2 +3 ; y=t^2+t
for which the tangent line passes through the origin.
I tried with 2t+3 and 2t+1
but it isn't successful

Answer 1

The slope at any point (x,y) is

dy/dx = (dy/dt)/(dx/dt) = (2t+1)/(2t)

We want to find t such that the tangent line is y=kx (that is, it passes through the origin). At that point

y = kx, so
k = (t^2+t)/(t^2+3)

That means that we need the slope = k, or

(2t+1)/(2t) = (t^2+t)/(t^2+3)
solve that and you get

t = 3+2√3

Check:
x = t^2+3 = 24+12√3
y = t^2+t = 24+14√3
dy/dx = 1/2 + 1/√3
y/x = 1/2 + 1/√3

so, y = kx, where k=1/2 + 1/√3