Parallel to which line because
yx = 0
is equ. of rectangular hyperbola

Given:
Hyperbola equation: x^2 - 3y^2 = 9
Tangent line: y = mx, where m is the slope of the line

To Find:
A point on the hyperbola where the tangent is parallel to the line y = mx

Explanation:
To find a point on the hyperbola where the tangent is parallel to y = mx, we need to determine the slope of the tangent line and compare it with the slope of y = mx.

Step 1: Find the derivative of the hyperbola equation:
Differentiating the given equation with respect to x will give us the slope of the tangent line at any point on the hyperbola.

Differentiating both sides of the equation x^2 - 3y^2 = 9 with respect to x:
2x - 6y * (dy/dx) = 0

Simplifying the above equation, we get:
2x = 6y * (dy/dx)

Step 2: Determine the slope of the tangent line:
The slope of the tangent line is given by dy/dx. From the equation obtained in step 1, we can calculate dy/dx as:

dy/dx = 2x / (6y)

Step 3: Compare slopes:
To find a point on the hyperbola where the tangent is parallel to y = mx, the slope of the tangent line should be equal to the slope of y = mx. Therefore, we need to equate the slopes:

dy/dx = 2x / (6y) = m

Step 4: Solve for x and y:
Substituting mx for dy/dx:

2x / (6y) = m

Cross-multiplying and simplifying, we get:
2x = 6my
x = 3my

Substituting x = 3my into the hyperbola equation:

(3my)^2 - 3y^2 = 9
9m^2y^2 - 3y^2 = 9
(9m^2 - 3) y^2 = 9
y^2 = 9 / (9m^2 - 3)
y = ±√(9 / (9m^2 - 3))

Step 5: Find the corresponding x-coordinates:
Substituting the value of y into x = 3my:

x = 3m * ±√(9 / (9m^2 - 3))
x = ±3√(9m^2 / (9m^2 - 3))

Conclusion:
The point(s) on the hyperbola x^2 - 3y^2 = 9, where the tangent is parallel to the line y = mx, are given by the coordinates (±3√(9m^2 / (9m^2 - 3)), ±√(9 / (9m^2 - 3))).