Eccentricity of the hyperbola: 4 x2 - 8 x - 5 y2 + 10 y = 21 is

To find the eccentricity of the hyperbola, we first need to rewrite the equation in standard form. The standard form of a hyperbola is (x-h)²/a² - (y-k)²/b² = 1 for a horizontal hyperbola and (y-k)²/a² - (x-h)²/b² = 1 for a vertical hyperbola.

The given equation is 4x² - 8x - 5y² + 10y = 21.

First, we group the x terms and y terms together:

4(x² - 2x) - 5(y² - 2y) = 21.

Next, we complete the square for the x terms and the y terms. To complete the square, we take half of the coefficient of x or y, square it, and add it to both sides of the equation.

For the x terms, half of -2 is -1, and (-1)² = 1. For the y terms, half of -2 is -1, and (-1)² = 1.

So, we add 1 to the x terms and subtract 1 from the y terms:

4[(x - 1)² - 1] - 5[(y - 1)² - 1] = 21.

Simplify to get:

4(x - 1)² - 4 - 5(y - 1)² + 5 = 21.

Combine like terms:

4(x - 1)² - 5(y - 1)² = 20.

Divide through by 20 to get the equation in standard form:

(x - 1)²/5 - (y - 1)²/4 = 1.

Now, we can see that a² = 5 and b² = 4. The eccentricity of a hyperbola is given by the formula e = sqrt(1 + (a²/b²)).

Substitute a² = 5 and b² = 4 into the formula to get:

e = sqrt(1 + (5/4)) = sqrt(2.25) = 1.5.

So, the eccentricity of the hyperbola is 1.5.