Find the equation of the normal to the curve `x^2/a^2-y^2/b^2=1` at `(x_0,y_0)`

Explanation

Given, `x^2/a^2-y^2/b^2=1` at `(x_0,y_0)`


Differentiating w.r.t `y`


`(2x)/(a^2)*(dx)/(dy)+(2y)/(b^2)=0`


`(dx)/(dy)=-(a^2y_1)/(b^2x_1)`



Line: `y=mx+c`


`y_1=(a^2y_1)/(b^2)+c`


`c=y_1(1-(a^2)/(b^2))`


`(y)/(y_1)=(a^2x)/(b^2)+(b^2-a^2)/(b^2)`


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