Find the equation of normal to the hyperbola `3x^2-y^2=1` having slope `1/3dot`

Explanation

Differentiating the equation of hyperbola `3x^(2)-y^(2)=1`
w.r.t.x, we get
`6x-2y(dy)/(dx)=0`
or `(dy)/(dx)=(3x)/(y)`
Let the point on the curve whereteh normal has slope 1/3 be `(x_(1),y_(1))`. Therefore,
`-(dx)/(dy)=-(y_(1))/(3x_(1))=(1)/(3)` or `1y_(1)=-x_(1) " " (1)`
Also, P lies on the curve. Therefore,
`3x_(1)^(2)-t_(1)^(2)=1 " " (2)`
Solving (1) and (2), we get
`x_(1)^(2)=(1)/(2)` or `x_(1)= pm(1)/(sqrt(2))`
`therefore y = pm (1)/(sqrt(2))`
Therefore, the points on the curve are `(pm1//sqrt(2),pm1//sqrt(2))`. Hence, the equations of normal are
`y-(1)/(sqrt(2))=(1)/(3)(x+(1)/(sqrt(2)))`
and `y+(1)/(sqrt(2))=(1)/(3)(x-(1)/(sqrt(2)))`
or `sqrt(2)(x-3y)=4` and `sqrt(2)(x-3y)=-4`
Alternative method :
Equation of normal to hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` havings lope m is
`y=mx pm((a^(2)+b^(2))m)/(sqrt(a^(2)-b^(2)m^(2)))`
So, for given hyperbola equation of normal having slope `(1)/(3)` is
`y=(1)/(3)x pm(((1)/(3)+1)(1)/(3))/(sqrt((1)/(3)-(1)/(9)))`
`rArr y=(1)/(3)x pm (2sqrt(2))/(3)`