Question

Find the asymptotes of the curve $2x^2 + 5xy + 2y^2 + 4x + 5y = 0$, and find the general equation of all hyperbola having the same asymptotes.

Answer 1

Asymptotes: $2x + y + 2 = 0$ and $x + 2y + 1 = 0$. General equation of hyperbolas: $2x^2 + 5xy + 2y^2 + 4x + 5y + C = 0$, where $C \neq 2$.

Answer 2

The given equation of the curve is $2x^2 + 5xy + 2y^2 + 4x + 5y = 0$. This is a general second-degree equation.

The equation of the pair of asymptotes is $2x^2 + 5xy + 2y^2 + 4x + 5y + k = 0$.

For this equation to represent a pair of straight lines, the determinant of the coefficients must be zero. Solving for $k$, we find that $k = 2$.

The equation of the pair of asymptotes is $2x^2 + 5xy + 2y^2 + 4x + 5y + 2 = 0$.

To find the individual asymptotes, we factor the equation. The quadratic part $2x^2 + 5xy + 2y^2$ factors as $(2x + y)(x + 2y)$.

We assume the equation of the pair of asymptotes is of the form $(2x + y + c_1)(x + 2y + c_2) = 0$. Expanding and comparing coefficients, we get:

$2c_2 + c_1 = 4$ $c_2 + 2c_1 = 5$ $c_1c_2 = 2$

Solving this system, we find $c_1 = 2$ and $c_2 = 1$.

The equations of the asymptotes are $2x + y + 2 = 0$ and $x + 2y + 1 = 0$.

The general equation of a hyperbola having the same asymptotes $L_1 = 0$ and $L_2 = 0$ is given by $L_1 L_2 = K$, where $K$ is a non-zero constant. So, $(2x + y + 2)(x + 2y + 1) = K$, where $K \neq 0$.

Expanding this gives $2x^2 + 5xy + 2y^2 + 4x + 5y + 2 = K$.

$2x^2 + 5xy + 2y^2 + 4x + 5y + (2 - K) = 0$.

Let $C = 2 - K$. Since $K \neq 0$, $C \neq 2$.

The general equation of all hyperbolas having the same asymptotes is $2x^2 + 5xy + 2y^2 + 4x + 5y + C = 0$, where $C$ is any constant except 2.