Finding the equation for a tangent line (Parametric Equation)

[PsychonautQQ]

Homework Statement

Find an equation of the tangent to the curve given by
x=tan(∅)
y=sec(∅)
at the point (1,sqrt(2))

The answer should be in the form y=f(x) without ∅

The Attempt at a Solution

Tangent line equation...
y-y*=m(x-x*)
m = dy/dx
m = sec(∅)tan(∅) / (sec^2(∅)
m = tan∅ / sec(∅)
m = sin∅

This is what I did so far, I'm a bit lost obviously

 

[eumyang]

Hint: use "θ" for theta, not "∅".

Find out what θ is when x = 1 and y = sqrt(2) (this is from the given point). Plug the value for θ in your expression for dy/dx to get your slope.

 

[PsychonautQQ]

Okay, so i did that and end up with m = sqrt(2)/2.
If i use the equation
y-y*=m(x-x*)
with
y* = sqrt(2)
x* = 1 and solve it i get
y=(sqrt(2)/2)x+2.1
but i type this into my online course and it says it's wrong?

 

[eumyang]

y=(sqrt(2)/2)x+2.1

I'm not getting 2.1.

 

[PsychonautQQ]

When i type in
y=.71x+.71 it gives me the wrong answer as well. ;-(

 

[eumyang]

Maybe it is because you are using a decimal approximation? I'm pretty sure the exact form of the last equation you wrote is correct.