# Equation of tangents of hyperbolas

1. Feb 9, 2013

### phospho

find the equation of the tangents to the hyperbola H` with equation $\frac{x^2}{25} - \frac{y^2}{16} = 1$ at the point (1,4)

in an earlier part of the equation we had to prove that a tangent to the a hyperbola in the form of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is in the form of $a^2m^2 = b^2 + c^2$ where the tangent is in the form of y = mx + c

so I differentiated with respect to x and got

$$\dfrac{dy}{dx} = \dfrac{16x}{25y}$$

subbing in the values of x and y I get the value of dy/dx to be 4/25

dy/dx is the gradient of the tangent so subbing that into the equation $a^2m^2 = b^2 + c^2$ as well as the values for a^2 and b^2 I can't get any real values for the constant, c so I'm not sure where I've gone wrong.

2. Feb 9, 2013

### SteamKing

Staff Emeritus
You have calculated m from evaluating dy/dx for the hyperbola at (1,4)

the equation of the tangent line is y = mx+c, so c (for the line) can be determined
by subbing in (1,4) for (x,y)

The relation a^2m^2 = b^2+c^2 is not that of a tangent line

3. Feb 9, 2013

### phospho

what is the relation of a^2m^2 = b^2+c^2 then?

4. Feb 9, 2013

### SteamKing

Staff Emeritus
5. Feb 9, 2013

### phospho

understand now, thank you.