In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

Tangents and Normals To Rectangular Hyperbolas

Go back to  'Hyperbola'

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 :

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

Download SOLVED Practice Questions of Tangents and Normals To Rectangular Hyperbolas for FREE
Hyperbolas
grade 11 | Questions Set 1
Hyperbolas
grade 11 | Answers Set 1
Hyperbolas
grade 11 | Questions Set 2
Hyperbolas
grade 11 | Answers Set 2
Download SOLVED Practice Questions of Tangents and Normals To Rectangular Hyperbolas for FREE
Hyperbolas
grade 11 | Questions Set 1
Hyperbolas
grade 11 | Answers Set 1
Hyperbolas
grade 11 | Questions Set 2
Hyperbolas
grade 11 | Answers Set 2
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school