Part (a): Deriving the equation for $x$-coordinates of intersection

Step 1: Substituting $y = 2x + k$ into the equation of the curve $C$

The equation of the curve is: $2x^2 + y^2 + 4x - 2y = 3$ Substitute $y = 2x + k$: $2x^2 + (2x + k)^2 + 4x - 2(2x + k) = 3$

Step 2: Expanding $(2x + k)^2$

$(2x + k)^2 = 4x^2 + 4kx + k^2$ So the equation becomes: $2x^2 + 4x^2 + 4kx + k^2 + 4x - 4x - 2k = 3$

Step 3: Simplifying the terms

Combine like terms: $6x^2 + 4kx + k^2 - 2k = 3$ Rearrange to form a standard quadratic equation: $6x^2 + 4kx + k^2 - 2k - 3 = 0$

This confirms the equation for the $x$-coordinates of the points where $C$ and $I$ intersect: $6x^2 + 4kx + k^2 - 2k - 3 = 0$

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Part (b): Finding the values of $k$ for which $I$ is a tangent to $C$

Step 1: Tangency condition

For the line $I$ to be tangent to the curve $C$, the quadratic equation in $x$ must have exactly one solution. This happens when the discriminant $\Delta$ of the quadratic equation is zero: $\Delta = b^2 - 4ac = 0$ Here, the quadratic equation is: $6x^2 + 4kx + (k^2 - 2k - 3) = 0$

• $a = 6$
• $b = 4k$
• $c = k^2 - 2k - 3$

Step 2: Calculate the discriminant

The discriminant is: $\Delta = (4k)^2 - 4(6)(k^2 - 2k - 3)$ Expand $\Delta$: $\Delta = 16k^2 - 24(k^2 - 2k - 3)$ Distribute $-24$: $\Delta = 16k^2 - 24k^2 + 48k + 72$ Simplify: $\Delta = -8k^2 + 48k + 72$

Step 3: Solve $\Delta = 0$

Set $\Delta = 0$: $-8k^2 + 48k + 72 = 0$ Divide through by $-8$ to simplify: $k^2 - 6k - 9 = 0$

Step 4: Solve the quadratic equation for $k$

Use the quadratic formula: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = 1$,
• $b = -6$,
• $c = -9$.

Substitute: $k = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-9)}}{2(1)}$ $k = \frac{6 \pm \sqrt{36 + 36}}{2}$ $k = \frac{6 \pm \sqrt{72}}{2}$ $k = \frac{6 \pm 6\sqrt{2}}{2}$ $k = 3 \pm 3\sqrt{2}$

Final Answer for Part (b):

The two values of $k$ are: $k = 3 + 3\sqrt{2} \quad \text{and} \quad k = 3 - 3\sqrt{2}$

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Summary:

(a) The quadratic equation for the $x$-coordinates of intersection is: $6x^2 + 4kx + k^2 - 2k - 3 = 0$

(b) The line $I$ is tangent to $C$ when $k = 3 + 3\sqrt{2}$ or $k = 3 - 3\sqrt{2}$.

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Would you like further clarification on any part?

Here are 5 related questions to deepen your understanding:

1. How do we verify that a line is tangent to a curve using derivatives?
2. What geometric interpretation does the discriminant provide for quadratic equations?
3. Can this approach be generalized to find tangents to other conic sections?
4. How does the $k$-value affect the position of the line $I$?
5. What is the significance of $\sqrt{2}$ in the solution for $k$?

Tip: When solving quadratic equations, always ensure the discriminant is simplified before proceeding with further calculations to avoid errors.