Here's the question you clicked on:
calyne
Find an equation of the tangent line to the hyperbola: x^2/a^2 - y^2/b^2 = 1 at the point (x0,x1)
implicit diff for this one
\[\frac{2x}{a^2}-\frac{2y}{b^2}y'=0\] \[y'=\frac{x}{a^2}\times \frac{b^2}{y}=\frac{b^2x}{a^2y}\]
yeah i got that but so what's the equation of the tangent line i don't get the textbook's answer
\[\frac{2x}{a^{2}}-\frac{2y}{b^{2}}y'=0\]\[y'=\frac{b^{2}x}{a^{2}y}\]\[m=\frac{b^{2}x_{0}}{a^{2}x_{1}}\]the equation of the tangent line is\[y-x_{1}=\frac{b^{2}x_{0}}{a^{2}x_{1}}\left( x-x_{0} \right)\]
yeah well the textbook says x0x/a^2 - y0y/b^2 = 1 so what's that about
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
\[a^{2}b^{2}=b^{2}x_{0}x-a^{2}x_{1}y\]\[\frac{x_{0}x}{a^{2}}-\frac{x_{1}y}{b^{2}}=1\]same as what the textbook said :)