Find the area of the region bounded by the hyperbola 9x² - 4y² = 36 and the line x = 3

Solution:

Given, the equation of hyperbola is 9x² - 4y² = 36

Given, the equation of line is x = 3.

We have to find the area of the region bounded by the given hyperbola and the line.

The standard form of the hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

Where, x is the major axis and y is the minor axis.

Converting the given hyperbola equation to standard form,

Dividing by 36 on both sides,

9x² - 4y² = 36 becomes x²/4 - y²/9 = 1

The vertex of the hyperbola is (±a, 0) when the major axis is the x-axis.

Here, a = ±2, b = ±3

On sketching the graph of the hyperbola,

The hyperbola is reflected about the x-axis so the area below equals the area above.

On rewriting the given equation of the hyperbola in terms of y,

9x² - 36 = 4y²

4y² = 9(x² - 4)

y² = (9/4)(x² - 4)

Taking square root,

y = (3/2)(√x² - 4)

Thus, the total area is double of the area of the region from x = 2 to x = 3.

Area = \(2\times \int_{2}^{3}\frac{3}{2}\sqrt{x^{2}-4}\, dx\)

Area = \(2\times \frac{3}{2}\int_{2}^{3}\sqrt{x^{2}-4}\, dx\)

Area = \(3\int_{2}^{3}\sqrt{x^{2}-4}\, dx\)

Using trigonometric substitution,

x = 2 sec(w)

dx = 2 sec(w) tan(w) dw

So, x² = 4sec²(w)

By using trigonometric identity,

sec²x - tan²x = 1

sec²x - 1 = tan²x

x² - 4 = 4sec²(w) - 4 = 4(sec²(w) - 1)

x² - 4 = 4 tan²(w)

On integrating,

\(\int \sqrt{x^{2}-4}\, dx=\int \sqrt{4tan^{2}(w)}(2sec(w)tan(w))dw\\=\int 2tan(w)(2sec(w)tan(w))dw\\=\int 4sec(w)tan^{2}(w)dw\)

On integrating by parts,

Let u = 4 tan(w)

du = 4 sec²(w) dw

Let v = sec(w)

dv = sec(w) tan(w) dw

\(\int 4sec(w)tan^{2}(w)dw=4sec(w)tan(w)-\int4sec(w)sec^{2}(w)dw\\\\=4sec(w)tan(w)-\int4sec(w)(tan^{2}(w)+1)dw\\\\=4sec(w)tan(w)-\int4sec(w)tan^{2}(w)dw-\int4sec(w)dw\\\int 4sec(w)tan^{2}(w)dw+\int4sec(w)tan^{2}(w)dw\\=4sec(w)tan(w)-\int 4sec(w)dw\)

\(\int 8sec(w)tan^{2}(w)dw=4sec(w)tan(w)-\int 4sec(w)dw\\\int 4sec(w)tan^{2}(w)dw=2sec(w)tan(w)-\int 2sec(w)dw\)

\(\int 4sec(w)tan^{2}(w)dw=2sec(w)tan(w)-2ln(sec(w)+tan(w))\)

Now, sec(w) = x/2

tan²(w) = sec²(w) - 1 = (x/2)² - 1

tan²(w) = (x² - 4)/4

Taking square root,

tan(w) = (√x² - 4)/2

So, \(\int \sqrt{x^{2}-4}=2sec(w)tan(w)-2ln(sec(w)+tan(w))\\\\=[2(\frac{x}{2})\frac{\sqrt{x^{2}-4}}{2}]-2ln[\frac{x}{2}+\frac{\sqrt{x^{2}-4}}{2}]\\\\=(\frac{x}{2}\sqrt{x^{2}-4})-2ln(\frac{x+\sqrt{x^{2}-4}}{2})\)

Applying limits,

Area = \(3[(\frac{3}{2}\sqrt{(3)^{2}-4})-2ln(\frac{3+\sqrt{(3)^{2}-4}}{2})\\-((\frac{2}{2}\sqrt{(2)^{2}-4})-2ln(\frac{2+\sqrt{(2)^{2}-4}}{2}))]\\\\=3[(\frac{3\sqrt{5}}{2}-2ln\frac{3+\sqrt{5}}{2})-(0)]\\\\=\frac{9\sqrt{5}}{2}-6ln(\frac{3+\sqrt{5}}{2})\)

Therefore, area of the region is \(A=\frac{9\sqrt{5}}{2}-6ln(\frac{3+\sqrt{5}}{2})\).

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Find the area of the region bounded by the hyperbola 9x² - 4y² = 36 and the line x = 3

Summary:

The area of the region bounded by the hyperbola 9x² - 4y² = 36 and the line x = 3 is \(A=\frac{9\sqrt{5}}{2}-6ln(\frac{3+\sqrt{5}}{2})\).