Mutually tangent line

Finding the area by integration is relatively straightforward once the equation of the tangent is known, so I will only explain in detail how I found the equation of the tangent and leave out the details of the integration.

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Let the tangent intersect the curve $y=-\frac {1} {x}$ at the point $(p,-\frac {1} {p})$ and the curve $y=\sqrt {x}$ at the point $(q, \sqrt {q})$ .

Since $\frac {dy} {dx} (-\frac {1} {x}) = \frac {1} {x^2}$ , the gradient of the tangent at $p$ is equal to $\frac {1} {p^2}$ , and the equation of the line in terms of $p$ is given by $y=\frac {1} {p^2} x -\frac {2} {p}$ (using $y=m(x-x_0)+y_0$ ).

Similarly, $\frac {dy} {dx} (\sqrt {x}) = \frac {1} {2\sqrt {x} }$ , so the gradient of the tangent at $q$ is equal to $\frac {1} {2\sqrt {q} }$ , and the equation of the line is $y= \frac {1} {2\sqrt {q} } x +\frac {\sqrt {q}} {2}$ .

Equating coefficients of $x$ and constants we find that

$\frac {1} {p^2} = \frac {1} {2\sqrt {q} }$ and $-\frac {2} {p} =\frac {\sqrt {q}} {2}$ . Substituting,

$\frac {1} {p^2} = \frac {1} {2(-\frac {4} {p}) }$

$\frac {1} {p^2} = -\frac {p} {8}$

$p^3=-8$ and hence $p=-2$ . Substituting this value into $y=\frac {1} {p^2} x -\frac {2} {p}$ we find that $y=\frac {1} {4} x+1$ .

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From here the area is found by evaluating $\int _{ -4 }^{ 0 }{ (\frac { 1 }{ 4 } x+1)dx } +\int _{ 0 }^{ 4 }{ (\frac { 1 }{ 4 } x+1-\sqrt { x } )dx } =\frac { 8 }{ 3 }$ .