To find the equation of the tangent to the circle at the point Q, we can use the fact that the radius at the point of tangency is perpendicular to the tangent. The slope of the radius is given by \( \frac{dy}{dx} \) at the point Q.
First, differentiate the equation of the circle \( x^2 + y^2 = 5 \) implicitly with respect to x:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Now, find \( \frac{dy}{dx} \) at the point Q (4/3, √29/3):
\[ 2\left(\frac{4}{3}\right) + 2\left(\frac{\sqrt{29}}{3}\right) \frac{dy}{dx} = 0 \]
Solve for \( \frac{dy}{dx} \), then use it to find the slope of the tangent. Finally, use the point-slope form of a line \( y - y_1 = m(x - x_1) \) with Q(4/3, √29/3) to determine the equation of the tangent.