Eliminate $t$ to give an equation that relates $x$ and $y$

Question

I am having problems understanding how to solve the following parametric equation. I have achieved an answer, but am unsure if my answer is correct or not.

Eliminate t to give an equation that relates x and y:

$x= \sec(t)$, $y=\tan^2(t)-2$

must solve for:

$y=$

My solution

$t = \sec^{-1}(x)$

Insert this value of $t$ in the equation for $y$.

$y=\tan^2(\sec^{-1}(x))-2$

Supplementary notes: for a point on the the curve:

$f(t)= \sec(t)$

$g(t)=\tan^2(t)-2$

where $z$ is some number

$P_{point}(f(z),g(z))$

Please confirm or reject my answer.

Answer 1

From "supplementary notes" onwards I don't quite understand. An easier way is using that $\tan^2t + 1 = \sec^2t$, so that: $$y = \tan^2t-2 = 1+\tan^2t - 3 = \sec^2t - 3 = x^2-3.$$

Answer 2

The identity $\color{blue}{\sec^2t=\tan^2t+1}$ imply $$y+2=\tan^2t=\sec^2 t-1=x^2-1$$

Answer 3

$\cos^{2} t+\sin^{2} t =1$, so $1+\tan^{2}t = \sec^{2}t$ so $y+3=x^{2}$

Answer 4

I will add another answer, because the method is so useful for any calculus problem involving trig substitutions or anything of the sort. (You will not need to remember any trig identities based on the Pythagorean Theorem.)

Starting off with $x = \sec t$, construct a right-triangle have the appropriate side lengths (here hypotenuse=$x$, adjacent=$1$) and determine the "other" side length using the Pythagorean Theorem.

Then simply read the values of other trig functions from the triangle by their definitions:

$\tan t = \dfrac{opposite}{adjacent} = \sqrt{x^2-1}$

so

$\tan^2 t = x^2-1$