The equation of normal to the curve 3x² – y² = 8 which is parallel to the line ,x + 3y = 8 is

The correct option is x + 3y ± 8 = 0.

Step-by-Step Explanation:

1. Find the slope of the given line:
The equation of the given line is:
$$x+3y=8$$
Rewriting this in slope-intercept form ($y=mx+c$), we get:
$$3y=-x+8$$
$$y=-\frac{1}{3}x+\frac{8}{3}$$
Thus, the slope of the given line is $m_1=-\frac{1}{3}$.

Since the normal to the curve is parallel to this line, the slope of the normal ($m_N$) must be equal to the slope of the line:
$$m_N=-\frac{1}{3}$$

2. Determine the slope of the tangent:
We know that the product of the slope of the tangent ($m_T$) and the slope of the normal ($m_N$) at any point on the curve is $-1$ ($m_T\cdot m_N=-1$).
Therefore:
$$m_T=-\frac{1}{m_N}=-\frac{1}{-\frac{1}{3}}=3$$

3. Find the points of contact on the curve:
The equation of the curve is given by:
$$3x^2-y^2=8$$
Differentiating both sides with respect to $x$ to find the rate of change (which represents the slope of the tangent, $\frac{dy}{dx}$):
$$\frac{d}{dx}(3x^2-y^2)=\frac{d}{dx}\left(8\right)$$
$$6x-2y\frac{dy}{dx}=0$$
$$2y\frac{dy}{dx}=6x$$
$$\frac{dy}{dx}=\frac{3x}{y}$$

Equating the derivative to the required slope of the tangent ($m_T=3$):
$$\frac{3x}{y}=3$$
$$x=y$$

Now, substitute $y=x$ back into the equation of the curve:
$$3x^2-x^2=8$$
$$2x^2=8$$
$$x^2=4$$
$$x=\pm 2$$

Since $y=x$, the coordinates of the points of contact are:
$(2,2)$ and $(-2,-2)$.

4. Find the equations of the normals:
Using the point-slope form of a line equation, $y-y_1=m_N(x-x_1)$, with slope $m_N=-\frac{1}{3}$:

Case 1: For the point $(2,2)$:
$$y-2=-\frac{1}{3}(x-2)$$
$$3(y-2)=-(x-2)$$
$$3y-6=-x+2$$
$$x+3y-8=0$$

Case 2: For the point $(-2,-2)$:
$$y-(-2)=-\frac{1}{3}(x-(-2\left)\right)$$
$$3(y+2)=-(x+2)$$
$$3y+6=-x-2$$
$$x+3y+8=0$$

Combining both cases, the equations of the normals are:
$$x+3y\pm 8=0$$