Find the equation(s) of normal(s) to the curve `3x^2-y^2=8` which is (are) parallel to the line `x+3y=4.`

Explanation

Given equation of the curve is
`3x^(2)-y^(2)=8`…………….(i)
On differentiating both sides w.r.t x, we get
`6x-2y(dy)/(dx) =0`
`rArr (dy)/(dx)=(6x)/(2y) = (3x)/y`
`rArr m_(1) = (3x)/y` [Say]
and slope of normal `(m_(2)) = -1/m_(1)=-y/(3x)` ..................(ii)
Since, slope of normal to the curve should be equal to the slope of line `x+3y=4`, which is parallel to curve.
For line, `y=(4-x)/(3) = -x/3+4/3`
`rArr` slope of the line `(m_(3)) = -1/3`
`therefore m_(2)=m_(3)`
`rArr -y/(3x) = -1/3`
`rArr -3y=-3x`
`rArr y=x`
On substituting the value of y in Eq. (i), we get
`3x^(2)-x^(2)=8`
`rArr x^(2)=4`
For x=2, `x=+-2`
and for x`=-2` `y=-2`
and for `x=-2` `y=-2` [using eq. (iii)]
Thus, the points at which normal to the curve are parallel to the line `x+3y=4` are `(2,2)` and `(-2,-2)`.
Required equations of normal are
`y-2=m_(2)(x-2)` and `y+2=m_(2)(x+2)`
`rArr y-2=-2/6(x-2)` and `y+2=-2/6(x+2)`
`rArr 3y-6=-x+2` and `3y+6=-x-2`
`rArr 3y+x=+8` and `3y+x=-8`
So,the required equations are `3y+x=+-8`