Let’s analyze and solve the problem step by step. The task involves finding the value of $k$ and the coordinates of the point of tangency.

Problem Setup

1. Line Equation: $x + 3y = k$
2. Curve Equation: $x^2 + y^2 + 2y - 9 = 0$

The condition for tangency means that the line touches the curve at exactly one point, so the system of equations must have exactly one solution.

Step 1: Substituting $y$ from the line equation into the curve equation

From the line equation: $y = \frac{k - x}{3}$ Substitute $y = \frac{k - x}{3}$ into the curve equation $x^2 + y^2 + 2y - 9 = 0$:

$x^2 + \left(\frac{k - x}{3}\right)^2 + 2\left(\frac{k - x}{3}\right) - 9 = 0$

Step 2: Simplify the equation

Expand $\left(\frac{k - x}{3}\right)^2$: $\left(\frac{k - x}{3}\right)^2 = \frac{(k - x)^2}{9} = \frac{k^2 - 2kx + x^2}{9}$

The equation becomes: $x^2 + \frac{k^2 - 2kx + x^2}{9} + \frac{2(k - x)}{3} - 9 = 0$

Multiply through by 9 to eliminate fractions: $9x^2 + (k^2 - 2kx + x^2) + 6(k - x) - 81 = 0$

Combine terms: $10x^2 - 2kx + k^2 + 6k - 6x - 81 = 0$

Step 3: Condition for tangency

For tangency, the quadratic equation in $x$: $10x^2 - (2k + 6)x + (k^2 + 6k - 81) = 0$ must have exactly one solution. This happens when the discriminant $\Delta$ is zero: $\Delta = B^2 - 4AC$ Here, $A = 10$, $B = -(2k + 6)$, and $C = k^2 + 6k - 81$.

Substitute into $\Delta = 0$: $(-(2k + 6))^2 - 4(10)(k^2 + 6k - 81) = 0$

Simplify: $(2k + 6)^2 - 40(k^2 + 6k - 81) = 0$

Expand $(2k + 6)^2$: $4k^2 + 24k + 36 - 40(k^2 + 6k - 81) = 0$

Distribute $-40$: $4k^2 + 24k + 36 - 40k^2 - 240k + 3240 = 0$

Combine terms: $-36k^2 - 216k + 3276 = 0$

Simplify by dividing through by -12: $3k^2 + 18k - 273 = 0$

Step 4: Solve for $k$

Divide through by 3: $k^2 + 6k - 91 = 0$

Factorize: $(k + 13)(k - 7) = 0$

Since $k$ is a positive constant: $k = 7$

Step 5: Find the point of tangency

Substitute $k = 7$ into the line equation: $x + 3y = 7 \quad \Rightarrow \quad y = \frac{7 - x}{3}$

Substitute $y = \frac{7 - x}{3}$ into the curve equation $x^2 + y^2 + 2y - 9 = 0$: $x^2 + \left(\frac{7 - x}{3}\right)^2 + 2\left(\frac{7 - x}{3}\right) - 9 = 0$

We already derived this equation earlier: $10x^2 - (2k + 6)x + (k^2 + 6k - 81) = 0$ Substitute $k = 7$: $10x^2 - 20x + (49 + 42 - 81) = 0$ $10x^2 - 20x + 10 = 0$

Divide through by 10: $x^2 - 2x + 1 = 0$

Factorize: $(x - 1)^2 = 0$

So: $x = 1$

Substitute $x = 1$ into $y = \frac{7 - x}{3}$: $y = \frac{7 - 1}{3} = 2$

Final Answer

• $k = 7$
• Point of tangency: $(1, 2)$

Would you like further clarification on any step?

Relative Questions:

1. How do you find the discriminant of a quadratic equation and its significance?
2. What is the geometric interpretation of tangency in the context of a line and a curve?
3. Why is substituting one equation into another effective for solving systems of equations?
4. How does the condition $\Delta = 0$ ensure a single solution?
5. Can this method be applied to other conic sections like ellipses or hyperbolas?

Tip:

Always check your final solution by substituting back into both the line and the curve equations to verify tangency!