# Find the equation of tangent for the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1? \begin{array}{1 1}(a)\;\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\normalsize =1\\(b)\;\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\normalsize =1\\(c)\;\large\frac{xx_1}{b^2}-\frac{yy_1}{a^2}\normalsize =1\\(d)\;\text{None of these}\end{array} ## 1 Answer Let us consider a point (x_1,y_1) at which we should find the tangent Hence slope at point (x_1,y_1) is \large\frac{2x}{a^2}-\frac{2yy_1}{b^2}$$=0$
$\large\frac{2x}{a^2}=\frac{yy_1}{b^2}$
At $(x_1,y_1)$
$\large\frac{x_1b^2}{y_1a^2}$$=y' Hence equation of tangent is (y-y_1)=\large\frac{x_1b^2}{y_1a^2}$$(x-x_1)$
$\large\frac{yy_1}{b^2}-\frac{y_1^2}{b^2}=\frac{xx_1}{a^2}-\frac{x_1^2}{a^2}$
$\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\normalsize =1$
Hence (a) is the correct answer.