Find the tangent line to the parametric curve at t = -pi/4?

Question

x = sec^2(t)-1

y = tan(t)

Pope · Answer

What, this again? In your previous question I showed that the parametric equations are equivalent to this relationship of x and y:

x = y²

The graph is a parabola. Its vertex is the origin, and its axis is the x-axis. Your earlier question concerned differentiation, but I would not use calculus here at all.

Let t = -π/4.
x = 1
y = -1

The point of tangency is (1, -1). By elementary properties of a parabola that tangent line meets the axis at (-1, 0). Those two points give you this line:

x + 2y + 1 = 0

az_lender · Answer

But, isn't sec^2(t) - 1 just equal to tan^2(t) ?
So isn't the curve just the same as x = y^2 ? ...with certain points missing...

At t = -pi/4, you have y = -1 and x = 1.
The slope dy/dx is the local value of dy/dx = 1/[2*sqrt(x)] = 1/2.
The equation of the line is
y = (1/2)x - 3/2.

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Find the tangent line to the parametric curve at t = -pi/4?

2 Answers