2023-M2-07

Let $x=2\sin \theta$, then $dx=2\cos\theta d\theta$. Hence, we have

$\begin{array}{rcl}
y & = & \dint \dfrac{k-3(2\sin\theta)}{\sqrt{4-(2\sin\theta)^2}} \times 2\cos \theta d\theta \\
y & = & \dint \dfrac{k-6\sin \theta}{2\cos \theta} \times 2\cos\theta d\theta \\
y & = & \dint (k-6\sin \theta)d\theta \\
y & = & k\theta +6\cos \theta +C \\
y & = & k\theta +6\sqrt{1-\sin^2\theta} +C \\
y & = & k\sin^{-1}\dfrac{x}{2} +6 \sqrt{1-\left(\dfrac{x}{2}\right)^2} +C \\
y & = & k\sin^{-1} \dfrac{x}{2} +3\sqrt{4-x^2} +C
\end{array}$

Since $\Gamma$ passes through the origin, then sub. $(0,0)$ into the above equation, we have

$\begin{array}{rcl}
0 & = & k\sin^{-1}\dfrac{0}{2} + 3\sqrt{4-(0)^2} +C \\
C & = & -6
\end{array}$

Therefore, the equation of $\Gamma$ is $y = k\sin^{-1} \dfrac{x}{2} +3\sqrt{4-x^2} -6$.

1. Since $\Gamma$ has a turning point, then there exist a point on $(-2, 2)$ such that $\dfrac{dy}{dx}=0$. We have

   $\begin{array}{rcl}
   \dfrac{k-3x}{\sqrt{4-x^2}} & = & 0 \\
   k & = & 3x
   \end{array}$

   Since $-2< x < 2$, then $-6< k < 6$.

2. $\begin{array}{rcl}
   \dfrac{dy}{dx} & = & \dfrac{k-3x}{\sqrt{4-x^2}} \\
   \dfrac{d^2y}{dx^2} & = & \dfrac{\sqrt{4-x^2}(-3)-(k-3x)\times \frac{1}{2}(4-x^2)^\frac{-1}{2}\times (-2x)}{(\sqrt{4-x^2})^2} \\
   \dfrac{d^2y}{dx^2} & = & \dfrac{-3(4-x^2)+x(k-3x)}{(4-x)^2\sqrt{4-x^2}} \\
   \dfrac{d^2y}{dx^2} & = & \dfrac{kx-12}{(4-x)^2\sqrt{4-x^2}}
   \end{array}$

   For the point of inflexion,

   $\begin{array}{rcl}
   \dfrac{d^2y}{dx^2} & = & 0 \\
   \dfrac{kx-12}{(4-x)^2\sqrt{4-x^2}} & = & 0 \\
   kx & = & 12
   \end{array}$

   However, by the result of (b)(i), $-6< k <6$ and $-2 < x <2$, i.e. $-12< kx < 12$.

   Therefore, $\dfrac{d^2y}{dx^2} \neq 0$ on $(-2,2)$.

   Also, it is obvious that $\dfrac{d^2y}{dx^2}$ is well defined on $(-2,2)$.

   Hence, $\Gamma$ does not have a point of inflexion on $(-2,2)$.