解联立方程组。 $\displaystyle \frac{x}{2} + \frac{y}{3} = 4$ (1) $\displaystyle 2xy = 45$ (2) 求常数 $k$ 的值，使得直线 $y = kx - \frac{1}{3}k^{2}$ 与曲线 $y = x^{2} - \frac{1}{3}x - 1$ 相切。 由此，求出使得直线 $y = kx - \frac{1}{3}k^{2}$ 与曲线 $y = x^{2} - \frac{1}{3}x - 1$ 相交于两个不同点的最大整数 $k$ 的值。 给出理由并说明，关于直线 $y = 9x - 27$ 和曲线 $y = x^{2} - 4$ 你能得出什么结论？

1. Clear the fractions in equation (1) and express $y$ in terms of $x$.

From (1),
$\frac{x}{2}+\frac{y}{3}=4$
Multiply by $6$:
$3x+2y=24$
$2y=24-3x$
$y=12-\frac{3}{2}x$

2. Substitute the expression for $y$ into the product equation.

Using (2),
$2xy=45$
Substitute $y=12-\frac{3}{2}x$:
$2x\left(12-\frac{3}{2}x\right)=45$
$24x-3x^2=45$

3. Form and solve the quadratic equation for $x$.

$24x-3x^2=45$
$3x^2-24x+45=0$
Divide by $3$:
$x^2-8x+15=0$
Factorise:
$(x-3)(x-5)=0$
Hence,
$x=3 \quad \text{or} \quad x=5$

4. Find the corresponding values of $y$ and give the solution pairs.

When $x=3$,
$y=12-\frac{3}{2}(3)=12-\frac{9}{2}=\frac{15}{2}$
When $x=5$,
$y=12-\frac{3}{2}(5)=12-\frac{15}{2}=\frac{9}{2}$
So the solutions are
$(x,y)=\left(3,\frac{15}{2}\right) \quad \text{or} \quad \left(5,\frac{9}{2}\right).$

1. Use the tangent boundary values to identify the integer values of $k$ giving two distinct intersections.

From part (b)(i), the tangent values are
$k=1-\frac{2\sqrt{30}}{3}\quad \text{and}\quad k=1+\frac{2\sqrt{30}}{3}.$
For two distinct points of intersection,
$1-\frac{2\sqrt{30}}{3}<k<1+\frac{2\sqrt{30}}{3}.$
Since
$1+\frac{2\sqrt{30}}{3}\approx 4.65,$
the largest integer value of $k$ is
$k=4.$

1. Form the quadratic equation for the points of intersection.

At the points of intersection,
$9x-27=x^2-4.$
So
$x^2-9x+23=0.$

2. Use the discriminant to decide whether there are real intersections.

For $x^2-9x+23=0$,
$b^2-4ac=(-9)^2-4(1)(23)=81-92=-11.$
Since $-11<0$, there are no real roots.