Find equation of the tangent to the curve $x^2 - y^2 = 1$ at the point $\sec \theta, \tan \theta$

Question

I am working through a pure maths book as a hobby and have come across this problem:

Find equation of the tangent to the curve $x^2 - y^2 = 1$ at the point $(\sec \theta, \tan \theta)$

I have said:

$y^2 = x^2 - 1$

$2y.\frac{dy}{dx} = 2x \implies \frac{dy}{dx} = \frac{x}{y}$

$\frac{y - \tan \theta}{x - \sec \theta} = \frac{x}{y}$

I am not sure about this last line since the equation involves $y^2$ and $x^2$

In any event, I cannot see how to arrive at the answer, which is given as $y = x\ cosec \theta - \cot \theta$

Answer 1

You are almost there.
$$\frac{dy}{dx} = \frac{x}{y}$$ Slope at given point $$\frac{\sec \theta}{\tan \theta}=\csc \theta$$ So the equation of tangent using point-slope form- $$(y-\tan \theta)=\csc \theta(x-\sec \theta)$$ $$y=x\csc \theta-\csc \theta\sec \theta+\tan \theta$$ $$-\csc \theta\sec \theta+\tan \theta=\frac{\sin \theta}{\cos \theta}-\frac{1}{\cos \theta\sin \theta}=\frac{\sin^2\theta-1}{\cos \theta\sin \theta}=\frac{-\cos^2\theta}{\cos \theta\sin \theta}=-\cot\theta$$ $$y=x\csc \theta-\cot\theta$$ That's the answer.

Answer 2

Just a tiny mistake.

You see, the equation of a tangent to a curve $y=f(x)$ at the point $P(x_1,y_1)$ is written as \begin{equation*} \frac{y-y_1}{x-x_1} =f'( x_1) \end{equation*} In your problem, \begin{equation*} ( x_1,y_1) \equiv (\sec \theta ,\tan \theta ) \end{equation*} Can you go on from here?