Key Concepts and Formulas

• Area under a curve: The area under the curve $y=f(x)$ from $x=a$ to $x=x$ is given by $\int_a^x f(t) \, dt$.
• Fundamental Theorem of Calculus: $\frac{d}{dx} \int_a^x f(t) \, dt = f(x)$.
• Normal to a curve: The slope of the normal to the curve $y=f(x)$ at a point $(x, y)$ is given by $-\frac{dx}{dy} = -\frac{1}{dy/dx}$.

Step-by-Step Solution

Step 1: Express $A_1$ in terms of $x$ and $y$

We are given that $A_1 = 2A_2$ and $A_1 + A_2 = xy - 8$. Substituting $A_2 = \frac{A_1}{2}$ into the second equation, we get: $A_1 + \frac{A_1}{2} = xy - 8$ $\frac{3A_1}{2} = xy - 8$ $A_1 = \frac{2}{3}(xy - 8)$

This expresses $A_1$ in terms of $x$ and $y$.

Step 2: Find the area $A_1$ using integration

$A_1$ represents the area under the curve $y=y(x)$ from some initial $x$ value (say $x_0$) to the current $x$ value. Thus, we can write: $A_1 = \int_{x_0}^x y(t) \, dt = \int_{x_0}^x y \, dx$

Step 3: Form the differential equation

Combining the expressions for $A_1$ from Steps 1 and 2, we have: $\int_{x_0}^x y \, dx = \frac{2}{3}(xy - 8)$

Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus, we get: $y = \frac{2}{3}\left(y + x \frac{dy}{dx}\right)$

Step 4: Solve the differential equation

Multiplying both sides by 3, we get: $3y = 2y + 2x \frac{dy}{dx}$ $y = 2x \frac{dy}{dx}$ Separating variables, we have: $\frac{dy}{y} = \frac{dx}{2x}$ Integrating both sides, we get: $\int \frac{dy}{y} = \int \frac{dx}{2x}$ $\ln|y| = \frac{1}{2} \ln|x| + C$ $\ln y = \ln \sqrt{x} + C$ (Since we are in the first quadrant, $x>0$ and $y>0$) $y = e^{\ln \sqrt{x} + C} = e^C e^{\ln \sqrt{x}} = e^C \sqrt{x}$ Let $e^C = k$, so $y = k\sqrt{x}$, where $k$ is a constant.

Step 5: Determine the value of $k$

When $x=x_0$, $A_1 = 0$. Substituting this into the equation $A_1 = \frac{2}{3}(xy - 8)$, we get: $0 = \frac{2}{3}(x_0 y_0 - 8)$ $x_0 y_0 = 8$ Since $y = k\sqrt{x}$, then $x_0 k\sqrt{x_0} = 8$. Also, $A_1 + A_2 = xy - 8$ and $A_1 = 2A_2$, so $3A_2 = xy - 8$. When $x=x_0$, $A_2 = 0$, so $x_0y_0 = 8$.

Substituting $y = k\sqrt{x}$ back into $A_1 = \frac{2}{3}(xy - 8)$, we get $\int_{x_0}^x k\sqrt{t} dt = \frac{2}{3}(x(k\sqrt{x}) - 8)$ $k \frac{2}{3}t^{3/2}|_{x_0}^x = \frac{2}{3}(kx^{3/2} - 8)$ $k(x^{3/2} - x_0^{3/2}) = kx^{3/2} - 8$ $-kx_0^{3/2} = -8$ $kx_0^{3/2} = 8$ Since $x_0y_0 = 8$ and $y_0 = k\sqrt{x_0}$, we have $x_0k\sqrt{x_0} = 8$, which means $kx_0^{3/2} = 8$.

Now, let's consider the case when $x=4$. Then $x_0y_0 = 8$ and $y_0 = k\sqrt{x_0}$. $y = k\sqrt{4} = 2k$. Also, $4(2k) = 8$, so $8k=8$, which means $k=1$. Therefore, $y = \sqrt{x}$. Then $x_0y_0 = 8$, so $x_0 \sqrt{x_0} = 8$, which gives $x_0 = 4$ and $y_0 = 2$.

Thus, $y = \sqrt{x}$.

Step 6: Find the equation of the normal

The slope of the tangent is $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$. The slope of the normal is $-\frac{dx}{dy} = -2\sqrt{x}$. We want the normal to be perpendicular to the line $2x - 12y = 15$, which has a slope of $\frac{2}{12} = \frac{1}{6}$. Therefore, the slope of the normal must be $-6$. So, $-2\sqrt{x} = -6$, which means $\sqrt{x} = 3$, and $x = 9$. Then $y = \sqrt{9} = 3$. The equation of the normal at $(9, 3)$ with slope $-6$ is: $y - 3 = -6(x - 9)$ $y - 3 = -6x + 54$ $y = -6x + 57$

Step 7: Check which point does NOT lie on the normal

Now we check which of the given points does not satisfy the equation $y = -6x + 57$:

• (A) (6, 21): $21 = -6(6) + 57 = -36 + 57 = 21$. This point lies on the normal.
• (B) (8, 9): $9 = -6(8) + 57 = -48 + 57 = 9$. This point lies on the normal.
• (C) (10, -4): $-4 = -6(10) + 57 = -60 + 57 = -3$. This point does NOT lie on the normal.
• (D) (12, -15): $-15 = -6(12) + 57 = -72 + 57 = -15$. This point lies on the normal.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when finding the slope of the normal.
• Integration Constants: Remember the constant of integration and use the initial conditions to find its value.
• Differentiation: Make sure to differentiate correctly using the chain rule where necessary.

Summary

We first found the area $A_1$ in terms of $x$ and $y$. Then, using the integral representation of $A_1$ and the Fundamental Theorem of Calculus, we derived a differential equation. Solving this differential equation gave us the equation of the curve $y=\sqrt{x}$. We then found the point on the curve where the normal has a slope of $-6$, giving us the point $(9, 3)$. The equation of the normal was then found to be $y = -6x + 57$. Finally, we checked which of the given points did not lie on this normal, finding that (10, -4) does not.

The final answer is \boxed{(10, -4)}, which corresponds to option (C).