Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the angle between the asymptotes of the hyperbola $\;$ $24x^2 - 8y^2 = 27$

Equation of hyperbola is $\;$ $24x^2 - 8y^2 = 27$

i.e. $\;$ $\dfrac{x^2}{27 / 24} - \dfrac{y^2}{27 / 8} = 1$

i.e. $\;$ $\dfrac{x^2}{9 / 8} - \dfrac{y^2}{27 / 8} = 1$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = \dfrac{9}{8}$ $\implies$ $a = \dfrac{3}{2 \sqrt{2}}$

and $b^2 = \dfrac{27}{8}$ $\implies$ $b = \dfrac{3 \sqrt{3}}{2 \sqrt{2}}$

The angle between the asymptotes is

$\begin{aligned} 2 \alpha & = 2 \tan^{-1} \left(\dfrac{b}{a}\right) \\\\ & = 2 \tan^{-1} \left(\dfrac{3 \sqrt{3} / 2 \sqrt{2}}{3 / 2 \sqrt{2}}\right) \\\\ & = 2 \tan^{-1} \left(\sqrt{3}\right) \end{aligned}$

i.e. $\;$ $2 \alpha = \dfrac{2 \pi}{3}$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the hyperbola if its asymptotes are parallel to $\;$ $x + 2y - 12 = 0$ $\;$ and $\;$ $x - 2y+ 8 = 0$, $\;$ $\left(2,4\right)$ is the center of the hyperbola and it passes through $\left(2,0\right)$.

Equations of the given straight lines are

$x + 2y - 12 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x - 2y + 8 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(1)$ is $= m_1 = \dfrac{-1}{2}$

Slope of equation $(2)$ is $= m_2 = \dfrac{1}{2}$

$\because$ $\;$ the asymptotes are parallel to the given lines [equations $(1)$ and $(2)$],

$\therefore$ $\;$ the slopes of the asymptotes are $\dfrac{1}{2}$ and $\dfrac{-1}{2}$

Let the equations of the asymptotes be

$y = \dfrac{1}{2}x + p$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;$ $y = -\dfrac{1}{2}x + q$ $\;\;\; \cdots \; (3b)$

where $p$ and $q$ are the Y intercepts.

Solving equations $(3a)$ and $(3b)$ simultaneously gives the point of intersection of the asymptotes.

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$

$\dfrac{1}{2}x + p = \dfrac{-1}{2}x + q$ $\implies$ $x = q - p$

Substituting the value of $x$ in equation $(3a)$ gives

$y = \dfrac{1}{2} \left(q - p\right) + p$ $\implies$ $y = \dfrac{1}{2} \left(p + q\right)$

The asymptotes intersect at the center.

Given: Center of hyperbola $= \left(2, 4\right)$

$\therefore$ $\;$ We have $\;$ $q - p = 2$ $\;\;\; \cdots \; (4a)$

and $\;$ $\dfrac{1}{2} \left(p + q\right) = 4$ $\implies$ $p + q = 8$ $\;\;\; \cdots \; (4b)$

Solving equations $(4a)$ and $(4b)$ simultaneously we have,

$2q = 10$ $\implies$ $q = 5$

Substituting the value of $q$ in equation $(4a)$ gives

$p = q - 2 = 3$

$\therefore$ $\;$ The equations of the asymptotes are

$y = \dfrac{1}{2}x + 3$ $\;\;$ i.e. $\;$ $x - 2y + 6 = 0$

and $\;$ $y = \dfrac{-1}{2}x + 5$ $\;\;$ i.e. $\;$ $x + 2y - 10 = 0$

$\therefore$ $\;$ The combined equations of the asymptotes is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) = 0$ $\;\;\; \cdots \; (5)$

The equation of the hyperbola differs from the combined equation of the asymptotes by a constant.

$\therefore$ $\;$ The equation of the hyperbola is of the form

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + k = 0$ $\;\;\; \cdots \; (6)$

The required hyperbola passes through the point $\left(2, 0\right)$.

$\therefore$ $\;$ We have from equation $(6)$,

$\left(2 - 0 + 6\right) \left(2 + 0 - 10\right) + k = 0$

$\implies$ $k = 64$

Substituting the value of $k$ in equation $(6)$, the equation of the required hyperbola is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + 64 = 0$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the asymptotes to the hyperbola $8x^2 + 10xy - 3y^2 - 2x + 4y -2 = 0$

Equation of given hyperbola is $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y - 2 = 0$ $\;\;\; \cdots \; (1)$

The combined equation of the asymptotes differs from the hyperbola by a constant only.

$\therefore$ $\;$ Let the combined equation of the asymptotes be: $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y + k = 0$

Now,

$\begin{aligned} 8x^2 + 10xy - 3y^2 & = 8x^2 + 12xy - 2xy - 3y^2 \\\\ & = 4x \left(2x + 3y\right) - y \left(2x + 3y\right) \\\\ & = \left(4x - y\right) \left(2x + 3y\right) \end{aligned}$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + \ell = 0$ $\;\;\; \cdots \; (2a)$ and

$2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $\left(4x - y + \ell\right) \left(2x + 3y + m\right) = 8x^2 + 10xy - 3y^2 - 2x + 4y + k$ $\;\;\; \cdots \; (3)$

Equating the coefficient of the $x$ terms we have

$4 m + 2 \ell = -2$

i.e. $\;$ $2 m + \ell = -1$ $\;\;\; \cdots \; (4a)$

Equating the coefficient of the $y$ terms we have

$- m + 3 \ell = 4$ $\;\;\; \cdots \; (4b)$

Equating the constant term we have

$\ell m = k$ $\;\;\; \cdots \; (4c)$

Solving equations $(4a)$ and $(4b)$ simultaneously we get

$\ell = 1$ $\;$ and $m = -1$

$\therefore$ $\;$ We have from equation $(4c)$, $\;$ $k = -1$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + 1 = 0$ $\;$ and $\;$ $2x + 3y - 1 = 0$

The combined equation of the asymptotes is

$8x^2 + 10xy - 3y^2 - 2x + 4y - 1 = 0$