Find your Math Personality!

Tangents to Hyperbolas


In this section, we discuss tangents in different forms for hyperbolas. We also discuss chords of contact, chords bisected at given points, pair of tangents from external points and so on, as we have already done for the case of other conic sections. The forms of the equations for all these will turn out to be the same as the results obtained for other conics.

We will use the hyperbola  \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) in our discussion.

TANGENT AT (x1, y1) : Consider a point \(P({x_1},{y_1})\) on the given hyperbola. The slope of the tangent at this point can be obtained by differentiating the equation of the hyperbola.

\[\begin{align}& \qquad \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\\\\&\Rightarrow \quad \frac{{2x}}{{{a^2}}} - \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0\\\\ &\Rightarrow \quad \frac{{dy}}{{dx}} = \frac{{{b^2}x}}{{{a^2}y}}\\ \\&\Rightarrow \quad {\left. {\frac{{dy}}{{dx}}} \right|_{P({x_1},{y_1})}} = \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}\end{align}\]

Thus, the equation of the tangent will be

\[\begin{align}&\qquad\quad\quad  y - {y_1} = \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}(x - {x_1}) \\ \\  &\Rightarrow\qquad   \frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} = \frac{{x_1^2}}{{{a^2}}} - \frac{{y_1^2}}{{{b^2}}} \\ \\  &\Rightarrow\qquad   \boxed{\frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} = 1}\,\,\,\,\,\,\,\,\,\,\left( \begin{gathered}  RHS{\text{ above is 1 since (}}{x_{\text{1}}},{y_1})\\   {\text{lies on the hyperbola}} \end{gathered}  \right)\end{align} \]

This equation is sometimes written more concisely as \(T({x_1},\,\,{y_1}) = 0\)

TANGENT AT  (a sec\(\theta \), b tan\(\theta \) ) : If the point P has been specified in parametric form, the equation for the tangent at P can be obtained by the substitution \({x_1} \to a\sec \theta \) and \({y_1} \to b\tan \theta \) in the equation for the tangent we obtained above :

\[\boxed{\frac{x}{a}\sec \theta  - \frac{y}{b}\tan \theta  = 1}\]

This is the equation of the tangent at the point \(\theta .\)

TANGENT OF SLOPE m : In Example -10 in the previous section, we’ve already obtained the equation of any tangent of slope m to the hyperbola \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1:\end{align}\)

\[\boxed{y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} }\]

The coordinates of the point(s) of contact, as already mentioned earlier, are

\[\left( { \pm \frac{{{a^2}m}}{{\sqrt {{a^2}{m^2} - {b^2}} }},\,\, \pm \frac{{{b^2}}}{{\sqrt {{a^2}{m^2} - {b^2}} }}} \right)\]

Example - 11

Show that two tangents can be drawn to a hyperbola from any point P lying outside the parabola.

Solution : Let the equation of the hyperbola be \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) and the coordinates of P be (h, k).

Any tangent of slope m to this hyperbola will have the equation

\[y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} \]

If this passes through \(P(h,\,\,k),\) we have

\[\begin{align}&\qquad\;\; k = mh \pm \sqrt {{a^2}{m^2} - {b^2}}  \\\\&\Rightarrow  \quad {(k - mh)^2} = {a^2}{m^2} - {b^2}\\\\ &\Rightarrow  \quad ({h^2} - {a^2}){m^2} - 2hkm + {k^2} + {b^2} = 0\end{align} \]

This is a quadratic in m which will give us two roots, corresponding to two tangents. These roots are real if the discriminant is positive, i.e., if

\[\begin{align}&\quad\;\;\;\;\;\;4{h^2}{k^2} > 4({h^2} - {a^2})({k^2} + {b^2}) \\\\&\Rightarrow  \quad {h^2}{b^2} < {a^2}{k^2} + {a^2}{b^2} \\\\&\Rightarrow  \quad \frac{{{h^2}}}{{{a^2}}} - \frac{{{k^2}}}{{{b^2}}} < 1 \\\\&\Rightarrow  \quad S(h,\,\,k) < 0\\\end{align} \]

Thus, two real tangents to the hyperbola can be drawn from any point only if that point lies outside the hyperbola. If a point lies on a hyperbola, only one tangent can be drawn (basically, coincident tangents) at the same point, while for any point inside the hyperbola, tangents cannot be drawn to the hyperbola.

Download SOLVED Practice Questions of Tangents to 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