Area Between Hyperbola and its Tangent Lines

Question

Draw tangent lines from point $\left(\frac{3}{5},0\right)$ on a hyperbola $x^2-y^2=1$ and then find the area defined by these tangent lines and a hyperbola.

Using some basic analytic geometry we can find that these lines are $t_1:y=\frac{-5}{4}x+\frac{3}{4}$ and $t_2:y=\frac{5}{4}x-\frac{3}{4}$. They intersect with hyperbola at points $x_1=\frac{5}{3}$ and $x_2=\frac{-5}{3}$. This is graphical representation:

We can see that part below $x$-axis is the same area as the one above it. So to find the total area we can just find the upper part and multiply it by $2$. So I did the following: $$P=2\left(\int_{\frac{3}{5}}^{\frac{5}{3}}{\frac{5}{4}x-\frac{3}{4}dx}-\int_{1}^{\frac{5}{3}}{\sqrt{x^2-1}dx}\right)$$ But this doesn't yield the correct solution, which is $\ln{3}-\frac{8}{9}\approx0.2097$. What am I doing wrong here? How to correctly do this? Can this be done using the form of equation in which $x=a\cosh{t}$ and $y=a\sinh{t}$ and if so, then how?

The problem is from the following book: https://etfuni.files.wordpress.com/2013/10/zbirka-zadataka-iz-vise-matematike-1-ilicic-uscumlic.pdf, problem number 3822

Answer 1

Considering the integrating domain as a x-simple set, we have:

$$ \begin{aligned} A & = 2\int_0^{\frac{4}{3}} \text{d}y \int_{\frac{4}{5}y+\frac{3}{5}}^{\sqrt{y^2+1}} \text{d}x \\ & = 2\left[\int_0^{\frac{4}{3}}\sqrt{y^2+1}\,\text{d}y - \int_0^{\frac{4}{3}}\left(\frac{4}{5}\,y+\frac{3}{5}\right)\text{d}y\right] \\ & = 2\left[\left(\frac{10}{9}+\frac{\log(3)}{2}\right)-\frac{68}{45}\right] \\ & = \boxed{\log(3) - \frac{4}{5}} \end{aligned} $$

while considering it as an y-simple set, we have:

$$ \begin{aligned} A & = \int_{\frac{3}{5}}^1 \text{d}x \int_{-\frac{5}{4}x+\frac{3}{4}}^{\frac{5}{4}x-\frac{3}{4}} \text{d}y + 2\int_1^{\frac{5}{3}} \text{d}x \int_{\sqrt{x^2-1}}^{\frac{5}{4}x-\frac{3}{4}} \text{d}y \\ & = 2\int_{\frac{3}{5}}^1 \left(\frac{5}{4}x-\frac{3}{4}\right)\text{d}x + 2\int_1^{\frac{5}{3}} \left(\frac{5}{4}x-\frac{3}{4}\right)\text{d}x - 2\int_1^{\frac{5}{3}} \sqrt{x^2-1}\,\text{d}x \\ & = 2\left[\int_{\frac{3}{5}}^{\frac{5}{3}} \left(\frac{5}{4}x-\frac{3}{4}\right)\text{d}x - \int_1^{\frac{5}{3}} \sqrt{x^2-1}\,\text{d}x\right] \\ & = 2\left[\frac{32}{45}-\left(\frac{10}{9}-\frac{\log(3)}{2}\right)\right] \\ & = \boxed{\log(3) - \frac{4}{5}}\,. \end{aligned} $$

It should be clear that the two approaches are equivalent and the result of the book is wrong.

Answer 2

It is unfortunate that you did not show how you carried out your evaluation of the integrals. You have for the area sought

$$ 2·\left( \ \int_{\frac35}^{\frac53} \ \left[ \ \frac54x \ - \ \frac34 \ \right] \ dx \ \ - \ \ \int_{1}^{\frac53} \ \sqrt{x^2-1} \ \ dx \right) \ \ . $$

The first integral represents the area of a right triangle with base $ \ \frac53 - \frac35 \ $ and height $ \ \frac43 \ \ , \ $ so its value is just $ \ \frac12·\left(\frac53 - \frac35 \right)·\frac43 \ = \ \frac{64}{90} \ = \ \frac{32}{45} \ \ . \ $ You are subtracting from that the area under the hyperbola, which may be found from a "secant substitution", $ \ x \ = \ \sec \theta \ \ : \ $

$$ \int_{1}^{\frac53} \ \sqrt{x^2-1} \ \ dx \ \ \rightarrow \ \ \int_{0}^{\Theta} \ \tan \theta \ \ ( \ \sec \theta \tan \theta \ \ d\theta \ ) \ \ = \ \ \int_{0}^{\Theta} \ \sec \theta \tan^2 \theta \ \ d\theta \ $$

[where $ \ \sec \Theta \ = \ \frac53 \ \ , \ \ \tan \Theta \ = \ \frac43 \ \ ] $

$$ = \ \ \int_{0}^{\Theta} \ \sec \theta · ( \ \sec^2 \theta \ - 1 \ ) \ \ d\theta \ \ = \ \ \int_{0}^{\Theta} \ ( \ \sec^3 \theta \ - \sec \theta \ ) \ \ d\theta $$

$$ = \ \ \frac12 · [ \ \tan \theta · \sec \theta \ + \ \ln ( \tan \theta \ + \ \sec \theta) \ \ + \ \ln ( \tan \theta \ + \ \sec \theta) \ ] \ \ - \ \ \ln ( \tan \theta \ + \ \sec \theta) \ \ \vert_0^{\Theta} $$

$$ = \ \ \frac12 · [ \ \tan \theta · \sec \theta \ - \ \ln ( \tan \theta \ + \ \sec \theta) \ \ ] \vert_0^{\Theta} $$

$$ = \ \ \frac12 \ · \ \left[ \ \frac43 · \frac53 \ - \ \ln \left( \frac43 \ + \ \frac53 \right) \ - \ 0·1 \ + \ \ln(0 \ + \ 1) \ \right] \ \ = \ \ \frac{10}{9} \ - \ \frac12 · \ln 3 \ \ . $$

The total area between the hyperbola and the specified tangent lines is then $$ 2·\left( \ \frac{32}{45} \ - \ \left[ \ \frac{10}{9} \ - \ \frac12 · \ln 3 \ \right] \ \right) \ \ = \ \ 2·\left( \ \frac{32 \ - \ 50}{45} \ + \ \frac12 · \ln 3 \ \right) \ \ = \ \ \ln 3 \ - \ \frac{36}{45} \ \ . $$

So if this is what you found, you are correct; the solver for the textbook apparently either got $ \ \frac{40}{45} \ $ or didn't simplify the fraction corrctly. (Around half-a-percent of answers in large course textbooks and problem collections, on average, are incorrect for various reasons, in my experience. Quite a few of the enquiries we get at MSE from students about problems are due to this. Upper-level undergraduate and graduate texts are rather more careful about such answers as they offer.)

This is the result πρόσεχε has shown. In answer to your question in the comment, they are using integration in the way it is carried out in multivariate calculus, showing that the same area is found in what corresponds to integration along the "$y-$direction" and along the "$x-$direction" in single-variable calculus.

To carry out the area "under" the hyperbola in parametric form, since the hyperbolic "Pythagorean Identity" is $ \ \cosh^2 \phi \ - \ \sinh^2 \phi \ = \ 1 \ \ , \ $ we would write $ \ x \ = \ \cosh \phi \ \ , \ \ y \ = \ \sinh \phi \ \ $ as you suggest, making the curve function $ \ y \ = \ \sqrt{\cosh^2 \phi - 1} \ = \ \sinh \phi \ \ , \ $ the differential $ \ dx \ = \ \sinh \phi \ d \phi \ \ , \ $ and the limits of integration $ \ a \ = \ 1 \ = \ \cosh 0 \ \Rightarrow \ \phi_{a} \ = \ 0 \ $ and $ \ b \ = \ \frac53 \ = \ \cosh \Phi \ \ $ with $ \ \sinh \Phi \ = \ \frac43 \ \ . \ $ We now have

$$ \int_{1}^{\frac53} \ \sqrt{x^2-1} \ \ dx \ \ \rightarrow \ \ \int_{0}^{\Phi} \ \sinh \phi \ \ ( \ \sinh \phi \ d \phi \ ) \ \ = \ \ \int_{0}^{\Phi} \ \sinh^2 \phi \ \ d \phi $$

$$ = \ \ \int_{0}^{\Phi} \ \frac12·(\cosh [2 \phi] \ - \ 1) \ \ d \phi \ \ = \ \ \frac12 · ( \ \frac12· \sinh [2 \phi] \ - \ \phi \ ) \vert_0^{\Phi} $$

[hyperbolic function identities and anti-derivatives are analogous to those for circular (trigonometric) functions]

$$ = \ \ \left( \ \frac14· 2·\sinh \phi · \cosh \phi \ - \ \frac12 ·\phi \ \right) \vert_0^{\Phi} $$ $$ = \ \ \ \frac12· \sinh \Phi · \cosh \Phi \ - \ \frac12 ·\cosh^{-1} \frac53 \ - \ \frac12· 0 · 1 \ - \ \frac12 ·0 $$

[ $ \ \cosh^{-1} x \ = \ \ln ( \ x \ + \ \sqrt{x^2 - 1} \ ) \ \ : \ $ seems like something we've done before? ]

$$ = \ \ \frac12· \frac43 · \frac53 \ - \ \frac12 ·\ln \left(\frac53 \ + \ \frac43 \right) \ \ = \ \ \frac{10}{9} \ - \ \frac12 ·\ln 3 \ \ , $$

giving us the same area we found above. (Many people here would have gone to the "hyperbolic substitution" for this integral immediately; the hyperbolic functions are frequently omitted in standard calculus courses, however, even when parametric curves are covered.)