Remove theta from these of trigonometric equations (basic)

Question

I have a very simple problem but am self-teaching and have become a bit stuck in finding an answer anywhere. The question is below and I have proceeded as follows:

'Please remove theta from the following pairs of equations:'

$ x = 4 \sec\theta\\\\ y = 4 \tan\theta$

$x = 4(1+\tan^2\theta)\\\\ y=4\tan\theta$

$x/4 = 1 + \tan^2\theta\\\\ y^2/4^4 = \tan^2\theta$

$x/4 - y^2/4^4 = 1$

Now in this situation I would always expand the square and multiply x/4 by 4/4 to get 4x/16, and my final answer was $4x-y^2 = 16$ however the book answer was $x^2 - y^2 = 16$. Are these equivalent (it doesn't seem like they are...) or have I gone wrong somewhere?

Many thanks for your help.

Answer 1

The third line is wrong. It should be

$$x^2 = 16(1 + tan^2\theta)$$

Then

$$\frac{x^2}{4^2} - \frac{y^2}{4^2} = 1$$

Which is the answer in the book.

Answer 2

You made an error in the third line:

$\dfrac x4=\sec \theta$, so $\sec^2\;\theta=1+\tan^2\theta=\dfrac{x^2}{16}$.

Can you proceed?

Answer 3

$$\begin{array} \\ x = 4 \tan \theta \\ y = 4 \sec \theta \end{array}$$ Dividing both equations by $4$ to isolate the trigonometric functions... $$\begin{array} \\ \dfrac {x} {4} = \tan \theta \\ \dfrac {y} {4} = \sec \theta \end{array}$$ Squaring both sides... $$\begin{array} \\ \dfrac {x^2} {16} = \tan^2 \theta \\ \dfrac {y^2} {16} = \sec^2 \theta \end{array}$$ Since $\tan^2 \theta - \sec^2 \theta = 1$ $$\dfrac {x^2} {16} - \dfrac {y^2} {16} = 1$$ and multiplying by $16$ gives us $$x^2 - y^2 = 16$$