Implicit Differentiation - Exercise 2

Exercise 2. Prove that an equation of the tangent line to the graph of the hyperbola

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

at the point P(x₀,y₀) is

$$\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1$$

Answer. Instead of finding y explicitly as a function of x, we will use implicit differentiation to find the slope of the tangent line. We have

$$2 \frac{x}{a^2} - \frac{2 y y'}{b^2} = 0$$

or equivalently

$$y' = \frac{x/a^2}{y/b^2} \;\cdot$$

So the equation of the tangent line at the point P(x₀,y₀) is

$$y - y_0 = \frac{x_0/a^2}{y_0/b^2} (x-x_0)$$

or equivalently

$$\frac{y_0}{b^2} (y - y_0) = \frac{x_0}{a^2} (x-x_0)\;.$$

Knowing that

$$\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1$$

the equation of the tangent line becomes

$$\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1\;.$$

[Back] [Next] [Trigonometry] [Calculus] [Geometry] [Algebra] [Differential Equations] [Complex Variables] [Matrix Algebra]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Copyright © 1999-2025 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour