## calyne Group Title Find an equation of the tangent line to the hyperbola: x^2/a^2 - y^2/b^2 = 1 at the point (x0,x1) 2 years ago 2 years ago

1. satellite73 Group Title

implicit diff for this one

2. satellite73 Group Title

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

3. calyne Group Title

yeah i got that but so what's the equation of the tangent line i don't get the textbook's answer

4. exraven Group Title

$\frac{2x}{a^{2}}-\frac{2y}{b^{2}}y'=0$$y'=\frac{b^{2}x}{a^{2}y}$$m=\frac{b^{2}x_{0}}{a^{2}x_{1}}$the equation of the tangent line is$y-x_{1}=\frac{b^{2}x_{0}}{a^{2}x_{1}}\left( x-x_{0} \right)$

5. calyne Group Title

yeah well the textbook says x0x/a^2 - y0y/b^2 = 1 so what's that about

6. exraven Group Title

algebra..$y-x_{1}=\frac{b^{2}x_{0}}{a^{2}x_{1}}\left( x-x_{0} \right)$$a^{2}x_{1}y-a^{2}x_{1}^{2}=b^{2}x_{0}x-b^{2}x_{0}^{2}$$b^{2}x_{0}^{2}-a^{2}x_{1}^{2}=b^{2}x_{0}x-a^{2}x_{1}y$look back at the hyperbola equation, plug the point (x0,x1)$\frac{x_{0}^{2}}{a^{2}}-\frac{x_{1}^{2}}{b^{2}}=1$$b^{2}x_{0}^{2}-a^{2}x_{1}^{2}=a^{2}b^{2}$substitute to the tangent line equation

7. exraven Group Title

$a^{2}b^{2}=b^{2}x_{0}x-a^{2}x_{1}y$$\frac{x_{0}x}{a^{2}}-\frac{x_{1}y}{b^{2}}=1$same as what the textbook said :)