# Finding tangent to parametric curves

• tnutty

#### tnutty

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.
x = tan(θ)
y = sec(θ)
(1 , √2)

y = ?

attempt ;

y - y1 = m(x-x1)

y = √2
x = 1

y1 = sec(θ)
x1 = tan(θ)

substituting and solving it out gives me,

√2 - sec(θ) = sin(θ)(1-tan(θ))

not sure how to solve for theta from there, even if i try to manipulate it.

You are taking y1 as sectheta and x1 as tantheta! That will never give any answer. Cause tantheta and sectheta represent any point on the curve.Where theta keeps on varying.So forget all that and start considering...

You have one point. And you have the equation. So replace x1 and y1 by 1 and root 2. But you still need to get the slope.

Slope is dy/dx. Now we need to get dy/dx. Its quite easy to find that? Try to find out. If you are unable to do so, ask us. We are always here to help you mate

sin(1)(x-1)+sqrt(2) is this right?

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Not really, that's not even an equation. Where's y? Try it by eliminating the parameter first. sec(t)^2-tan(t)^2=1, right? Where t is theta?

dy/dx = sin(theta)

I guess FedEx method does not work in this case because if I
use x1= 1 and y1 = sqrt(2), and use the point slope formula , then my last post does not work then.

So how could I find this?

dy/dx = sin(theta)

I guess FedEx method does not work in this case because if I
use x1= 1 and y1 = sqrt(2), and use the point slope formula , then my last post does not work then.

So how could I find this?

At some point you are going to have to find a value of theta that corresponds to x=1 and y=sqrt(2). Or find a way to derive sin(theta) from tan(theta) and sec(theta). You can't put x=1 in as a value of theta.

At some point you are going to have to find a value of theta that corresponds to x=1 and y=sqrt(2).

Precisely

Precisely

Or show tan(theta)/sec(theta)=sin(theta). They both work.

Ofcourse

dy/dx = sin(o). I got this from "Or show tan(theta)/sec(theta)=sin(theta). "

Now I can't use (1,sqrt(2)) as x1 and y1 with point slope formula , and use sin(1) as the slope?

Yes, use (1,sqrt(2)) as x1, y1. For the last time, no! Not sin(1). The slope is sin(THETA). Not sin(x).