To solve the problem step by step:

Question (b):

We are tasked to show that the $x$-coordinate of the intersection of the line $l$ with the curve $C$ satisfies the equation:

$2x^2 - 4x + k^2 = 0$

1. Curve Equation:

$y = \frac{k^2}{x} + 1$

2. Line Equation:

$y = -2x + 5$

3. Intersection of $C$ and $l$:

Equating the $y$-values: $\frac{k^2}{x} + 1 = -2x + 5$

Simplify: $\frac{k^2}{x} = -2x + 4$

Multiply through by $x$ (valid since $x \neq 0$): $k^2 = -2x^2 + 4x$

Rearranging gives: $2x^2 - 4x + k^2 = 0$

This verifies the required equation.

Question (c):

We now find the values of $k$ for which the line $l$ is tangent to the curve $C$. This means the quadratic equation:

$2x^2 - 4x + k^2 = 0$

has exactly one solution for $x$. For a quadratic to have one solution, its discriminant must equal zero.

Discriminant:

The general quadratic equation $ax^2 + bx + c = 0$ has discriminant: $\Delta = b^2 - 4ac$

For our equation $2x^2 - 4x + k^2 = 0$:

• $a = 2$,
• $b = -4$,
• $c = k^2$.

The discriminant is: $\Delta = (-4)^2 - 4(2)(k^2)$ $\Delta = 16 - 8k^2$

Set $\Delta = 0$ for tangency: $16 - 8k^2 = 0$

Solve for $k^2$: $8k^2 = 16 \quad \Rightarrow \quad k^2 = 2$

Thus, $k = \pm\sqrt{2}$.

Final Answers:

• (b) The $x$-coordinate of the intersection satisfies $2x^2 - 4x + k^2 = 0$.
• (c) The values of $k$ for which $l$ is tangent to $C$ are: $k = \pm\sqrt{2}.$

Let me know if you have further questions or want additional details!

Related Questions:

1. What happens when $k = 0$? Is the line $l$ still tangent to the curve?
2. What is the geometric interpretation of the discriminant in this scenario?
3. How do the roots of the quadratic equation change as $k$ varies?
4. Can the curve $C$ intersect the line $l$ at more than two points? Why or why not?
5. How would the solution change if the line $l$ had a different slope?

Tip:

Always check the discriminant when analyzing tangency conditions for quadratics, as it determines the number of real solutions (intersection points).