2.0 Standard Equation of Hyperbola

  Hyperbola

Let $S$ be the focus and $ZM$ the directrix of the hyperbola. Draw $SZ$ perpendicular to $ZM$. Divide $SZ$ internally and externally in the ratio $e:1\ (e>1)$ and Let $A$ and $A'$ be their internal and external point of division. Then, $$\frac{{SA}}{{AZ}} = e...(1)$$ and $$\frac{{SA'}}{{A'Z}} = e...(2)$$

Let $AA'=2a$ and take $C$ the mid-point of $AA'$ as origin. $$\therefore CA = CA' = a$$

Let $P(x,y)$ be any point on the hyperbola and $CA$ as $X$-axis, the line through $C$ perpendicular to $CA$ as $Y$-axis. Then adding $(1)$ and $(2)$, we get

$$\begin{equation} \begin{aligned} SA + SA' = e(AZ + A'Z) \\ CS - CA + CS + CA' = e(AA') \\ 2CS = e(2a){\text{ }}(\because CA = CA') \\ CS = ae \\\end{aligned} \end{equation} $$

$\therefore $ The focus $S$ is $(CS,0)$ i.e., $(ae,0)$.

Subtracting $(1)$ from $(2)$, we get

$$\begin{equation} \begin{aligned} SA - SA' = e(A'Z - AZ) \\ AA' = e[(CA' + CZ) - (CA - CZ)] \\ AA' = e(2CZ) \\ 2a = e(2CZ) \\ CZ = \frac{a}{e} \\\end{aligned} \end{equation} $$

$\therefore $ The directrix $MZ$ is $x = CZ = \frac{a}{e}$

$$x - \frac{a}{e} = 0{\text{ (}}\because {\text{e > 1, }}\therefore \frac{a}{e} < 1)$$

Now, Draw $PM$ perpendicular to $MZ$,

$$\begin{equation} \begin{aligned} \therefore \frac{{SP}}{{PM}} = e{\text{ (by definition of hyperbola)}} \\ SP = e \times PM \\\end{aligned} \end{equation} $$

Squaring both sides, we get

$$\begin{equation} \begin{aligned} S{P^2} = {e^2}P{M^2} \\ {(x - ae)^2} + {(y - 0)^2} = {e^2}{(x - \frac{a}{e})^2} \\ {(x - ae)^2} + {y^2} = {(ex - a)^2} \\ {x^2} + {a^2}{e^2} - 2aex + {y^2} = {e^2}{x^2} - 2aex + {a^2} \\ {x^2}({e^2} - 1) - {y^2} = {a^2}({e^2} - 1) \\ \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{a^2}({e^2} - 1)}} = 1 \\ \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1{\text{ where }}{b^2} = {a^2}({e^2} - 1) \\\end{aligned} \end{equation} $$

Generally, the equation of the hyperbola whose focus is the point $(h,k)$ and directrix is $lx+my+n=0$ and whose eccentricity is $e$, is$${(x - h)^2} + {(y - k)^2} = {e^2}\frac{{{{(lx + my + n)}^2}}}{{{l^2} + {m^2}}}$$