What is the maximal area between hyperbola and chord $AB$?

Question

Suppose $A$ and $B$ are variable points on upper branch of hyperbola $\mathcal{H}:\;x^2-y^2=-4$ such that $AB = 1$. What is the maximal area between $AB$ and $\mathcal{H}$?

Clearly $y= \sqrt{x^2+4}$. If we set $A\big(t,\sqrt{t^2+4}\big)$ then $B\big(t+h, \sqrt{(t+h)^2+4} \big)$ for some positive $h$, such that $$h^2 +\Big(\sqrt{(t+h)^2+4} -\sqrt{t^2+4} \Big)^2=1$$ and we have to find a minimum value of $$f(t) = \int_{t}^{t+h}\sqrt{x^2+4}\;dx$$

Notice that if $$F(x):= \int\sqrt{x^2+4}\;dx$$ then $$ F(x)= 2\ln\left(\left|\sqrt{x^2+4}+x\right|\right)+x\sqrt{\dfrac{x^2}{4}+1}$$ and things get complicated very quickly.

Answer 1

Parametrize the hyperbola as $x=2\sinh t $, $y= 2\cosh t$ and let the endpoints of the chord be $A(2\sinh a, 2\cosh a)$, $B(2\sinh b, 2\cosh b)$. Then, the equation of chord $AB$ is

$$y= x\tanh\frac{a+b}2 +2\frac{\cosh \frac{a-b}2}{\cosh\frac{a+b}2 }$$ and the area between the chord and hyperbola is \begin{align} K =&\int_{x_a}^{x_b}( y - \sqrt{x^2+4})dx\\ =&\int_a^b \left(2\tanh\frac{a+b}2 \sinh t + 2\frac{\cosh \frac{a-b}2}{\cosh\frac{a+b}2 } - 2 \cosh t \right)2\cosh t\>dt\\ =& 2\tanh\frac{a+b}2 (\sinh^2 b -\sinh^2a) +2\sinh(b-a)(2-\cosh (a+b) )-2(b-a) \end{align} Moreover, $1=AB^2=(x_b-x_a)^2 + (y_b-y_a)^2 $ leads to $$\cosh2b +\cosh 2a -2 \cos(a+b)=\frac14$$ $$\frac{d b}{da} =\frac{\sinh2a}{\sinh(a+b)-\sinh2b}$$ Then, it is straightforward to verify that $\frac{d b}{da} = 1 $, $\frac{d K}{da} = 0 $ if $b=-a$, which leads to $\sinh b =-\sinh a= \frac14$ and the maximum area $$K_{max}= \frac{\sqrt{17}}4-4\sinh^{-1}\frac14$$

Answer 2

Disclaimer: I am not sure since I am still new to calculus.

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Find the minimum value of $f(t) = \displaystyle\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx$.

The fundamental theorem of calculus can be applied here:

\begin{align*}\dfrac{d}{dt}f(t) &= \dfrac{d}{dt}\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx \\ f'(t) &=\left(\sqrt{(t + h)^{2} + 4}\right)\left(\dfrac{d}{dt}(t + h)\right) - \left(\sqrt{t^{2} + 4}\right)\left(\dfrac{d}{dt}(t)\right) \\f'(t) &= \left(\sqrt{(t + h)^{2} + 4}\right) - \left(\sqrt{t^{2} + 4}\right)\end{align*}

Then, set $f'(t) = 0$, \begin{align*}f'(t) &= 0 \\ \sqrt{(t + h)^{2} + 4} - \sqrt{t^{2} + 4} &= 0 \\ \sqrt{(t + h)^{2} + 4} &= \sqrt{t^{2} + 4} \\ (t + h)^{2} + 4 &=t^{2} + 4 \\ t^{2} + 2th + h^{2} + 4 &= t^{2} + 4 \\ 2th + h^{2} &= 0 \\ 2t &= -h \\ t&= -\dfrac{h}{2}\end{align*}

I don't know how to proceed now, but I hope that this can help.

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I used the variant of the FTOC which is: $$\dfrac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt = f(v(x))v'(x) - f(u(x))u'(x)$$

Answer 3

Here is a solution based on a geometrical result that can be seen either on the second figure here or on the first figure of this answer of mine. It deals with the area "swept" by a moving point on a rectangular hyperbola (in close connection with the analogous second figure for circular trigonometry).

We are going to work on an equivalent problem on a rotated and scaled version of your rectangular hyperbola, with equation

$$x^2-y^2=1.$$

Its right branch is parameterizable in this way:

$$x=\cosh t, \ \ \ y=\sinh t, \ \ \ t \in (-\infty,\infty).$$

Take a look at figure 1: curvilinear triangles $OIA$ and $OIB$ have resp. signed areas $a/2$, $b/2$. Their unsigned areas are $|a|/2$ and $|b|/2$. In the case of this figure, we have $a>0, \ b<0$.

The area of curvilinear triangle $OB(I)A$ (red area) is

$$\dfrac12(a-b)\tag{1}$$

(proof in my answer referenced upwards).

Fig. 1. A case where $a >0$, and $ \ b <0$ (other sign cases are possible).

Besides, the (signed) area of triangle $OBA$ is

$$\dfrac12 \det(\vec{OB},\vec{OA})=\dfrac12 \begin{vmatrix}\cosh b&\cosh a\\\sinh b &\sinh a\end{vmatrix}=\dfrac12 \sinh(a-b),\tag{2}$$

Therefore, the area of "lunule" $IBA$ (in blue) is equal to the difference (2)-(1):

$$f(a,b):=\dfrac12 \sinh(a-b)-\dfrac12 (a-b) \tag{3}$$

Remark: formula (3) shows that the lunule's area depends only on the difference $a-b$.

Now, let us take into account condition $AB=1$ under the form :

$$g(a,b):=(\cosh a-\cosh b)^2+(\sinh a-\sinh b)^2-1=0\tag{4}$$

Let us express that we want an extremum of $f$ (as given by (3)) under constraint (4), by expressing it under the existence of a Lagrange multiplier $\lambda$ such that

$$\begin{cases}\lambda \ \dfrac{\partial f(a,b)}{\partial a} &=& \dfrac{\partial g(a,b)}{\partial a} \\ \lambda \dfrac{\partial f(a,b)}{\partial b} &=& \dfrac{\partial g(a,b)}{\partial b} \end{cases}\tag{5}$$

giving:

$$\left\{\begin{array}{rr}\lambda (\cosh(a-b)-1) &=& 2(\cosh a - \cosh b)\sinh a + 2(\sinh a - \sinh b) \cosh a \\ \lambda (-\cosh(a-b)+1) &=& -2(\cosh a - \cosh b)\sinh b -2(\sinh a - \sinh b) \cosh b\end{array}\right.\tag{6}$$

Adding these two equations:

$$0=2 (\cosh a - \cosh b)(\sinh a - \sinh b)\tag{7}$$

$\cosh a = \cosh b$ gives $b=\pm a$, and $\sinh a = \sinh b$ gives $a=b$.

Solution $a=b$ isn't acceptable (flat triangle!). Therefore:

$$b=-a\tag{8}$$

As a consequence, the area-maximizing chord $AB$ is vertical.

Now, plugging (8) into (4) gives:

$$0+(2 \sinh a)^2=1 \ \ \ \ \implies \ \ \ \ a=\sinh^{-1}(\tfrac14)\tag{9}$$

(of course we could have taken the opposite value for $a$).

Using (9), we deduce that the common abscissa of $A$ and $B$ is:

$$\cosh(a)=\sqrt{1+\sinh(a)^2}=\dfrac14\sqrt{17}\tag{10}$$

Using (9) in (3), the area of lunule $IBA$ is:

$$\text{area}=\dfrac12 \sinh(2a)-a = \sinh(a)\cosh(a)-a$$

$$\text{area}=\tfrac{1}{16}\sqrt{17}-\sinh^{-1}(\tfrac14)$$

which has to be multiplied by $2^2$ to find back the result of Quanto, due to the initial downscaling $2 \to 1$ operated on the hyperbola.