# Tangents and Normals To Rectangular Hyperbolas

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We now discuss the equations of tangents and normal (in various forms) to a rectangular hyperbola that has been specified using its asymptotes as the coordinate axes, i.e., that has the equation $$xy={{c}^{2}}.$$

TANGENT AT P(x1, y1): The slope of the tangent at P can be obtained by differentiating the equation of the hyperbola :

\begin{align} &\qquad\;\;\; y=\frac{{{c}^{2}}}{x} \\ & \Rightarrow\quad \frac{dy}{dx}=-\frac{{{c}^{2}}}{{{x}^{2}}} \\ & {{\left. \Rightarrow\quad \frac{dy}{dx} \right|}_{P({{x}_{1}},{{y}_{1}})}}=-\frac{{{c}^{2}}}{x_{1}^{2}} \\\end{align}

Thus, the equation of the tangent at P is

\begin{align}&\quad\quad y - {y_1} = - \frac{{{c^2}}}{{x_1^2}}(x - {x_1}) \\ &\Rightarrow\quad y - {y_1} = \frac{{ - {y_1}}}{{{x_1}}}(x - {x_1})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\text{Using }}{y_1} = \frac{{{c^2}}}{{{x_1}}}} \right\} \\ &\Rightarrow\quad x{y_1} + y{x_1} = 2{x_1}{y_1} \\ & \Rightarrow\quad \boxed{x{y_1} + y{x_1} = 2{c^2}} \\&\qquad\qquad \qquad \text{or}\\&\qquad\quad\boxed{\frac{x}{{{x_1}}} + \frac{y}{{{y_1}}} = 2}\ \end{align}

TANGENT AT   P\left( {ct,\,\begin{align}\frac{c}{t}\end{align}} \right)   : We use the substitution  $${x_1} \to ct$$ and  {y_1} \to \begin{align}\frac{c}{t}\end{align}  in the equations obtained above.

Thus, the tangent at ‘t’ has the equation

$\boxed{\frac{x}{t} + yt = 2c}$

$\text {or}$

$\boxed{x + y{t^2} = 2ct}$

From this, it follows that the point of intersection of the tangents drawn at  $$'{t_1}'$$ and  $$'{t_2}'$$ will be

$\left( {\frac{{2c{t_1}{t_2}}}{{{t_1} + {t_2}}},\,\frac{{2c}}{{{t_1} + {t_2}}}} \right)$

Observe that the coordinates of this intersection point are respective harmonic means between the coordinates of the points at which the tangents have been drawn.

NORMAL AT P(x1, y1): The slope of the normal at P will be

{{m}_{N}}=\frac{-1}{\underset{\begin{align} & \,\,\,\,\,\,\,\,\downarrow \\ & \text{slope of} \\ & \text{tangent at} \\ & P \\ \end{align}}{\mathop{{{m}_{T}}}}\,}=\frac{x_{1}^{2}}{{{c}^{2}}}=\frac{{{x}_{1}}}{{{y}_{1}}}\qquad\qquad\qquad \left( \text{Using }{{x}_{1}}{{y}_{1}}={{c}^{2}} \right)

Thus, the equation of the normal is

\begin{align}&\qquad \;\;y - {y_1} = {m_N}(x - {x_1})\\&\Rightarrow \quad y - {y_1} = \frac{{{x_1}}}{{{y_1}}}(x - {x_1})\\&\Rightarrow \quad {\boxed{x{x_1} - y{y_1} = x_1^2 - y_1^2}}\end{align}

NORMAL AT   P\left( {ct,\,\begin{align}\frac{c}{t}\end{align}} \right): To obtain the normal at the point  $$'t',$$  use the substitution  {x_1} \to ct,\,\,{y_1} \to \begin{align}\frac{c}{t}\end{align} in the equation

for the normal obtained above :

\begin{align}&\qquad\;\;\;xct - \frac{{yc}}{t} = {c^2}{t^2} - \frac{{{c^2}}}{{{t^2}}} \\ &\Rightarrow\quad \boxed{x{t^3} - yt - c{t^4} + c = 0} \\ \end{align}