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.`

Options

A. `3X-Y=8`
B. `3X+Y+8=0`
C. `X+3Y+-8=0`
D. `X+3Y=0`

Answer

C

Explanation

We have, the equation of the curve is `3x^(2)-y^(2)=8`
Also, the given equation of the line is `x+3y=8`
`rArr 3y=8-x`
`rArr y=-x/3+8/3`
Thus, the slope of the line is `-1/3` which should be equal to slope of the equation of normal to the curve.
On differentating Eq. (i) w.r.t. x, we get
`6x-2y(dy)/(dx)=0`
`rArr (dy)/(dx)=(6x)/(2y)=(3x)/(y)="Slope of the curve"`
Now, slope of normal to the curve `=-1/(dy)/(dx)`
`=-1/(3x)/(y)=-y/(3x)`
`therefore -(y/(3x))=-1/3`
`rArr -3y=-3x`
`rArr y=x`
On substituting the value of the given equation of the curve, we get,
`3x^(2)-x^(2)=8`
`rArr x^(2)=8/2`
`rArr x=+-2`
For x=2, `3(2)^(2)-y^(2)=8`
`rArr y^(2)=4`
`rArr y=+-2`
and for `x=-2, 3(-2)^(2)-y^(2)=8`
`rArr y=+-2`
So, the points at which normals are parallel to the given line are `(+2, +-2)`.
Hence, the equation of normal at `(+-2,+-2)` is
`=y-(+-2)-1=-1/3[x-(+2)]`
`rArr 3[y-(+-2)]=-[x-(+-2)]`
`x+3y+-8=0`