# Finding the equation for a tangent line (Parametric Equation)

1. Sep 5, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
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 ∅

3. 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

2. Sep 5, 2013

### 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.

3. Sep 5, 2013

### 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?

4. Sep 5, 2013

### eumyang

I'm not getting 2.1.

5. Sep 5, 2013

### PsychonautQQ

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

6. Sep 5, 2013

### 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.