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Step-by-Step Solution
Step 1: Rewrite the given ellipse in standard form
The ellipse given is
$$
\frac{x^2}{3} + \frac{y^2}{4} = 4.
$$
Divide both sides by 4 to express it in the form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $:
$$
\frac{x^2}{3 \cdot 4} + \frac{y^2}{4 \cdot 4} = 1
\quad \Longrightarrow \quad
\frac{x^2}{12} + \frac{y^2}{16} = 1.
$$
Step 2: Identify the ellipse parameters
In $ \frac{x^2}{12} + \frac{y^2}{16} = 1 $, the semi-major axis is along the $y$-direction with
$$
a^2 = 16 \quad \Longrightarrow \quad a = 4,
$$
and the semi-minor axis is along the $x$-direction with
$$
b^2 = 12 \quad \Longrightarrow \quad b = \sqrt{12} = 2\sqrt{3}.
$$
The eccentricity of the ellipse is
$$
e_\text{ellipse} = \sqrt{1 - \frac{b^2}{a^2}}
= \sqrt{1 - \frac{12}{16}}
= \sqrt{\frac{4}{16}}
= \frac{1}{2}.
$$
Step 3: Determine the foci of the ellipse
For this ellipse, because $a$ is along the $y$-axis, the foci are at
$$
(0, \pm c), \quad \text{where } c = ae_\text{ellipse} = 4 \times \frac{1}{2} = 2.
$$
Hence, the ellipse’s foci are
$$
(0,2) \quad \text{and} \quad (0,-2).
$$
Step 4: Set up the hyperbola
We are told the hyperbola has its transverse axis along the major axis of the ellipse and has vertices at the ellipse’s foci.
Thus, the hyperbola’s vertices are
$$
(0,2) \quad \text{and} \quad (0,-2).
$$
Since the key axis is vertical (along $y$), the hyperbola in standard form is:
$$
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1,
$$
where the hyperbola’s “vertices” correspond to $y = \pm a$, so $a = 2 \implies a^2 = 4.$
Step 5: Use the hyperbola’s eccentricity to find $b^2$
The eccentricity of the hyperbola is given as $ e_h = \frac{3}{2}. $ Recall for a hyperbola of the form
$$
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1,
$$
the eccentricity satisfies
$$
e_h^2 = 1 + \frac{b^2}{a^2}.
$$
Therefore,
$$
\left(\frac{3}{2}\right)^2 = 1 + \frac{b^2}{4}.
$$
That gives
$$
\frac{9}{4} = 1 + \frac{b^2}{4}
\quad \Longrightarrow \quad
\frac{b^2}{4} = \frac{9}{4} - 1
= \frac{5}{4}
\quad \Longrightarrow \quad
b^2 = 5.
$$
Step 6: Write the hyperbola’s equation
The standard form, with $a^2=4$ and $b^2=5$, is
$$
\frac{y^2}{4} - \frac{x^2}{5} = 1.
$$
A convenient rearrangement is
$$
\frac{x^2}{5} - \frac{y^2}{4} = -1.
$$
Step 7: Check the given points and find which does not satisfy
We test each point in the equation
$$
\frac{x^2}{5} - \frac{y^2}{4} = -1.
$$
• For the point $ (5, 2\sqrt{3}) $:
$$
\frac{5^2}{5} - \frac{(2\sqrt{3})^2}{4}
= \frac{25}{5} - \frac{4 \cdot 3}{4}
= 5 - 3
= 2,
$$
which is not equal to $-1.$ Hence, $ (5,2\sqrt{3}) $ does not lie on the hyperbola.
Conclusion
The point
$$
\bigl(5,\,2\sqrt{3}\bigr)
$$
does not satisfy the hyperbola’s equation and thus does not lie on it.
