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

calculusimplicit-differentiationparametrictangent line

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 $y = x\ cosec \theta – \cot \theta$

Best Answer

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?

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