Find one of the areas bounded by the curves: (x²)/20 +

Find one of the areas bounded by the curves: $\frac{x^2}{20} + \frac{y^2}{5} = 1$, $y = \frac{4}{x}$.

We will locate the first-quadrant intersection points and integrate the gap between the two curves.

Step 1: Understand the curves

Ellipse: $\frac{x^2}{20}+\frac{y^2}{5}=1$, with semi-axes $a=2\sqrt5$ (along $x$) and $b=\sqrt5$ (along $y$).

Hyperbola: $y=\frac4x$, a rectangular hyperbola.

Step 2: Find intersection points

Substituting $y=\frac4x$ into the ellipse equation:

\[\frac{x^2}{20} + \frac{16}{5x^2} = 1\]

Multiplying by $20x^2$:

\[x^4 + 64 = 20x^2 \implies x^4 - 20x^2 + 64 = 0\]

Let $u=x^2$: $u^2-20u+64=0 \implies u=\frac{20\pm\sqrt{400-256}}{2}=\frac{20\pm12}{2}$, so $u=16$ or $u=4$.

Therefore $x=\pm4$ or $x=\pm2$. In the first quadrant: $(2,2)$ and $(4,1)$.

Step 3: Verify these points lie on the ellipse

At $(2,2)$: $\frac{4}{20}+\frac{4}{5}=\frac15+\frac45=1$ ✓. At $(4,1)$: $\frac{16}{20}+\frac15=\frac45+\frac15=1$ ✓.

Step 4: Determine which curve is above

At $x=3$: ellipse gives $y^2=5\left(1-\frac{9}{20}\right)=\frac{11}{4}$, so $y=\frac{\sqrt{11}}{2}\approx1.658$; hyperbola gives $y=\frac43\approx1.333$. Since $1.658>1.333$, the ellipse is above the hyperbola for $x\in(2,4)$.

Step 5: Set up the integral

\[A = \int_2^4 \left[\sqrt{5-\frac{x^2}{4}} - \frac{4}{x}\right] dx = \frac12\int_2^4\sqrt{20-x^2}\,dx - \int_2^4\frac4x\,dx\]

Step 6: Evaluate the ellipse's arc integral

Using $\int\sqrt{a^2-x^2}\,dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\arcsin\frac{x}{a}+C$ with $a^2=20$:

\[\frac12\int_2^4\sqrt{20-x^2}\,dx = \frac12\left[\frac{x\sqrt{20-x^2}}{2}+10\arcsin\frac{x}{2\sqrt5}\right]_2^4\]

At $x=4$: $\frac{4\cdot2}{2}+10\arcsin\frac{2}{\sqrt5}=4+10\arcsin\frac{2}{\sqrt5}$. At $x=2$: $\frac{2\cdot4}{2}+10\arcsin\frac{1}{\sqrt5}=4+10\arcsin\frac{1}{\sqrt5}$.

\[\frac12\int_2^4\sqrt{20-x^2}\,dx = \frac12\left[10\arcsin\frac{2}{\sqrt5}-10\arcsin\frac{1}{\sqrt5}\right] = 5\left[\arcsin\frac{2}{\sqrt5}-\arcsin\frac{1}{\sqrt5}\right]\]

Step 7: Evaluate the hyperbola integral

\[\int_2^4 \frac4x\,dx = 4[\ln x]_2^4 = 4(\ln4-\ln2) = 4\ln2\]

Step 8: Calculate the area

\[A = 5\left[\arcsin\frac{2}{\sqrt5}-\arcsin\frac{1}{\sqrt5}\right] - 4\ln 2\]

\[\boxed{A = 5\left(\arcsin\frac{2}{\sqrt5}-\arcsin\frac{1}{\sqrt5}\right) - 4\ln 2 \text{ square units}}\]

Step 9: Numerical value

Using $\arcsin\frac{2}{\sqrt5}\approx1.1071$ rad and $\arcsin\frac{1}{\sqrt5}\approx0.4636$ rad:

\[A \approx 5(1.1071-0.4636) - 4(0.6931) \approx 3.2175 - 2.7726 \approx 0.445 \text{ square units}\]

Summary

\[\boxed{A = 5\left(\arcsin\frac{2}{\sqrt5}-\arcsin\frac{1}{\sqrt5}\right) - 4\ln 2 \approx 0.445 \text{ square units}}\]

Key insight: Where an ellipse and a rectangular hyperbola cross, the lens-shaped area between them mixes an inverse-trigonometric term (from the elliptical arc) with a logarithmic term (from the hyperbola) — a common signature of areas bounded by conics of different eccentricity.