Introduction

This article delves into the mathematical problem where we need to find the values of (k) for which the line (x 2y k) is tangent to the curve (xy 20). Tangency implies a unique point of intersection between the line and the curve, where the line touches the curve without crossing it.

Step 1: Expressing (y) in Terms of (x)

Given the line equation:

(x 2y k)
Solving for (y):
(2y k - x)
(y frac{k - x}{2})

Step 2: Substituting (y) into the Curve Equation

Substitute (y frac{k - x}{2}) into the curve equation (xy 20):
(x left(frac{k - x}{2}right) 20)
(frac{xk - x^2}{2} 20)
(xk - x^2 40)
(x^2 - xk 40 0)

Step 3: Formulating a Quadratic Equation

We now have a quadratic equation in the form (ax^2 bx c 0):
(x^2 - kx 40 0)

Step 4: Determining the Condition for Tangency

For the line to be tangent to the curve, the quadratic equation must have exactly one solution. This occurs when the discriminant (D) is zero. The discriminant (D) for a quadratic equation (ax^2 bx c 0) is given by:

(D b^2 - 4ac)

In our equation, (a 1), (b -k), and (c 40). Thus:

(D (-k)^2 - 4 cdot 1 cdot 40 k^2 - 160)

Solving for (k) when (D 0):
(k^2 - 160 0)
(k^2 160)
(k pm sqrt{160} pm 4sqrt{10})

Conclusion

Therefore, the values of (k) for which the line is tangent to the curve are (k 4sqrt{10}) and (k -4sqrt{10}).

Note: This method effectively uses algebra and the properties of tangency to find the specific values of (k). The values (k 4sqrt{10}) and (k -4sqrt{10}) ensure that the line (x 2y k) is tangent to the curve (xy 20).