# Find the equation of normal for the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1 in parametric form? \begin{array}{1 1}(a)\;\large\frac{by}{\tan \theta}+\frac{ax}{\sec\theta}\normalsize =a^2+b^2\\(b)\;\large\frac{by}{\tan \theta}-\frac{ax}{\sec\theta}\normalsize =a^2+b^2\\(c)\;\large\frac{by}{\tan \theta}+\frac{ax}{\sec\theta}\normalsize =a^2-b^2\\(d)\;\text{None of these}\end{array} ## 1 Answer Equation of normal at (x_1,y_1) is \large\frac{b^2y}{y_1}+\frac{a^2x}{x_1}$$=a^2+b^2$
Now replacing $(x_1,y_1)$ by $(a\sec\theta,b\tan\theta)$
Hence $\large\frac{b^2y}{b\tan\theta}+\frac{a^2x}{a\sec\theta}$$=a^2+b^2 \large\frac{by}{\tan\theta}+\frac{ax}{\sec\theta}$$=a^2+b^2$
Hence (a) is the correct answer.