1. Fundamental Concepts: The Hyperbola and its Standard Equation

A hyperbola is defined as the locus of all points in a plane such that the absolute difference of their distances from two fixed points (called foci) is a constant. This constant difference is equal to $2a$, where $a$ is the length of the semi-transverse axis.

For a hyperbola centered at the origin $(0,0)$, its equation and properties depend on whether its transverse axis (the line segment connecting the vertices and passing through the foci) is horizontal (along the x-axis) or vertical (along the y-axis).

In this problem, the given vertices and focus lie on the x-axis, which immediately tells us it is a horizontal hyperbola centered at the origin. Its standard equation and key properties are:

• Standard Equation: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
• Center: $(0,0)$
• Vertices: $(\pm a, 0)$
• Foci: $(\pm ae, 0)$, where $e$ is the eccentricity.
• Relationship between $a, b, e$: $b^2 = a^2(e^2 - 1)$
• Eccentricity: For any hyperbola, $e > 1$.

Our primary objective is to determine the specific equation of the hyperbola using the provided information. Once we have the equation, we will test each given point to find the one that does not satisfy it.

2. Step-by-Step Derivation of the Hyperbola's Equation

We are provided with the coordinates of the vertices and one focus. We will use these to systematically find the values of $a$, $e$, and $b^2$, which are essential for constructing the hyperbola's equation.

Step 1: Determine the Center and Orientation of the Hyperbola.

• Given Information: The vertices are at $(-2, 0)$ and $(2, 0)$.
• Why this step is taken: The center of a hyperbola is always the midpoint of its vertices. The alignment of the vertices (both having a y-coordinate of 0) directly tells us the orientation of the transverse axis.
• Calculation: The midpoint of $(-2, 0)$ and $(2, 0)$ is $\left( \frac{-2+2}{2}, \frac{0+0}{2} \right) = (0,0)$. Since the vertices lie on the x-axis, the transverse axis is along the x-axis.
• Result: The hyperbola is centered at the origin $(0,0)$ and is a horizontal hyperbola.

Step 2: Determine the Value of 'a' (Semi-transverse Axis Length).

• Given Information: Vertices are $(\pm 2, 0)$.
• Why this step is taken: For a horizontal hyperbola centered at the origin, the vertices are defined as $(\pm a, 0)$. By comparing the given coordinates with this general form, we can directly find the value of $a$, which represents the distance from the center to each vertex.
• Calculation: Comparing $(\pm a, 0)$ with $(\pm 2, 0)$, we find $a = 2$.
• Result: Therefore, $a^2 = 2^2 = 4$.

Step 3: Determine the Value of 'ae' (Distance from Center to Focus).

• Given Information: One of the foci is at $(-3, 0)$.
• Why this step is taken: For a horizontal hyperbola centered at the origin, the foci are defined as $(\pm ae, 0)$. This directly gives us the value of $ae$, which is the distance from the center to each focus.
• Calculation: Comparing $(\pm ae, 0)$ with $(\pm 3, 0)$, we get $ae = 3$.
• Result: $ae = 3$.

Step 4: Calculate the Eccentricity 'e'.

• Known Values: We have $a = 2$ (from Step 2) and $ae = 3$ (from Step 3).
• Why this step is taken: Eccentricity ($e$) is a crucial parameter that describes the shape or "openness" of the hyperbola. We can determine it by combining the values of $a$ and $ae$. It also serves as an important check: for a hyperbola, $e$ must always be greater than $1$.
• Calculation: Substitute $a=2$ into the equation $ae=3$: $2e = 3$ $e = \frac{3}{2}$
• Verification: Since $e = 1.5$, which is indeed greater than $1$, our calculations are consistent with the properties of a hyperbola.

Step 5: Calculate the Value of 'b^2' (Semi-conjugate Axis Squared).

• Known Values: We have $a = 2$ and $e = \frac{3}{2}$.
• Why this step is taken: The parameter $b^2$ (or $b$) is necessary to complete the standard equation of the hyperbola. It is related to the semi-conjugate axis. The fundamental relationship $b^2 = a^2(e^2 - 1)$ connects $a$, $e$, and $b^2$.
• Calculation: $b^2 = a^2(e^2 - 1)$ Substitute $a = 2$ and $e = \frac{3}{2}$: $b^2 = (2)^2 \left( \left(\frac{3}{2}\right)^2 - 1 \right)$ $b^2 = 4 \left( \frac{9}{4} - 1 \right)$ $b^2 = 4 \left( \frac{9 - 4}{4} \right)$ $b^2 = 4 \left( \frac{5}{4} \right)$ $b^2 = 5$
• Result: $b^2 = 5$.

Step 6: Formulate the Standard Equation of the Hyperbola.

• Known Values: We found $a^2 = 4$ (from Step 2) and $b^2 = 5$ (from Step 5).
• Why this step is taken: This is the culmination of our parameter calculations. By substituting $a^2$ and $b^2$ into the standard equation for a horizontal hyperbola, we obtain the unique equation that defines the hyperbola described in the problem.
• Calculation: Substitute $a^2 = 4$ and $b^2 = 5$ into the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $\frac{x^2}{4} - \frac{y^2}{5} = 1$
• Conclusion: This is the required equation of the hyperbola.

3. Verifying Which Point Does Not Lie on the Hyperbola

Now that we have the equation of the hyperbola, $\frac{x^2}{4} - \frac{y^2}{5} = 1$, we need to check each given option. A point $(x,y)$ lies on the hyperbola if, when its coordinates are substituted into the equation, the Left Hand Side (LHS) evaluates to $1$ (the Right Hand Side, RHS). We are looking for the point that does not satisfy this condition.

Option (A): $\left( {6,5\sqrt 2 } \right)$ Substitute $x = 6$ and $y = 5\sqrt{2}$ into the hyperbola equation: $\text{LHS} = \frac{(6)^2}{4} - \frac{(5\sqrt{2})^2}{5}$ $= \frac{36}{4} - \frac{25 \times 2}{5}$ $= 9 - \frac{50}{5}$ $= 9 - 10$ $= -1$ Since $\text{LHS} = -1 \neq 1$, the point $\left( {6,5\sqrt 2 } \right)$ does not lie on the hyperbola. This is our answer. Let's verify the other options for completeness.

Option (B): $\left( {2\sqrt 6 ,5} \right)$ Substitute $x = 2\sqrt{6}$ and $y = 5$ into the hyperbola equation: $\text{LHS} = \frac{(2\sqrt{6})^2}{4} - \frac{(5)^2}{5}$ $= \frac{4 \times 6}{4} - \frac{25}{5}$ $= 6 - 5$ $= 1$ Since $\text{LHS} = 1 = 1$, the point $\left( {2\sqrt 6 ,5} \right)$ lies on the hyperbola.

Option (C): $\left( { - 6,2\sqrt {10} } \right)$ Substitute $x = -6$ and $y = 2\sqrt{10}$ into the hyperbola equation: $\text{LHS} = \frac{(-6)^2}{4} - \frac{(2\sqrt{10})^2}{5}$ $= \frac{36}{4} - \frac{4 \times 10}{5}$ $= 9 - \frac{40}{5}$ $= 9 - 8$ $= 1$ Since $\text{LHS} = 1 = 1$, the point $\left( { - 6,2\sqrt {10} } \right)$ lies on the hyperbola.

Option (D): $\left( {4,\sqrt {15} } \right)$ Substitute $x = 4$ and $y = \sqrt{15}$ into the hyperbola equation: $\text{LHS} = \frac{(4)^2}{4} - \frac{(\sqrt{15})^2}{5}$ $= \frac{16}{4} - \frac{15}{5}$ $= 4 - 3$ $= 1$ Since $\text{LHS} = 1 = 1$, the point $\left( {4,\sqrt {15} } \right)$ lies on the hyperbola.

As confirmed by the detailed checks, only Option (A) results in the LHS not equaling 1, meaning it does not lie on the hyperbola.

4. Important Tips and Common Mistakes to Avoid

• Identify Transverse Axis Correctly: Always start by determining if the transverse axis is horizontal or vertical. This is crucial for selecting the correct standard equation ($\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$). The coordinates of the vertices or foci will guide you.
• **Distinguish 'a' and 'b