Find the equation of normal to the hyperbola `x^2-9y^2=7` at point (4, 1).

Explanation

Differentiating the equation of hyperbola `x^(2)-9y^(2)=7a` w.r.t. x, we get
`2x-18y(dy)/(dx)=0`
`"or "(dy)/(dx)=(x)/(9y)`
slope of normal at any point one the curve is `-(dx)/(dy)=-(9y)/(x).`
Therefore, the slope of noraml at point (4, 1) is
`(-(dx)/(dy))_(("4,1"))=-(9)/(4)`
Therefore, the equation of normal at point (4, 1) is `y-1=-(9)/(4)(x-4)` `"or "9x+4y=40`
Alternative method,
For hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, equation of normal at point `P(x_(a),y_(1))` is
`(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))==a^(2)+b^(2)`
So, for given hyperbola `(x^(2))/(7)-(y^(2))/(7//9)=1`, equation of normal at point P(4, 1) is
`(7x)/(4)+((7//9)y)/(1)=7+(7)/(9)`
`rArr" "9x+4y=40`