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Step-by-Step Solution
Step 1: Convert the ellipse to its standard form
The given ellipse is:
$3x^2 + 4y^2 = 12.$
Dividing both sides by 12, we get:
$\frac{x^2}{4} + \frac{y^2}{3} = 1.$
Hence, the ellipse is in the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
with $a^2=4$ and $b^2=3.$
Step 2: Determine the eccentricity and foci of the ellipse
For an ellipse of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
(assuming $a^2 > b^2$), the eccentricity $e$ is:
$e = \sqrt{1 - \frac{b^2}{a^2}}.$
Here, $a^2 = 4 \implies a = 2$ and $b^2 = 3 \implies b = \sqrt{3}.$
So
$e = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}.$
The foci of the ellipse (since it is aligned along the x-axis) are
$(\pm ae, 0) = (\pm (2 \cdot \tfrac12), 0) = (\pm 1, 0).$
Step 3: Identify the given hyperbola parameters
We are told the hyperbola has the same foci as the ellipse,
and its transverse axis length is $\sqrt{2}.$
For a hyperbola of the form
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$
the transverse axis length is $2a.$
Since $2a = \sqrt{2},$ we get $a = \frac{1}{\sqrt{2}}.$
Step 4: Relate hyperbola’s foci to its eccentricity
The eccentricity of a hyperbola
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
is
$e = \sqrt{1 + \frac{b^2}{a^2}}.$
The coordinates of its foci are
$(\pm ae,\, 0) = \left(\pm a\sqrt{1 + \frac{b^2}{a^2}},\, 0\right).$
From Step 2, the common foci with the ellipse are $(\pm 1, 0).$
Hence,
$a \sqrt{1 + \frac{b^2}{a^2}} = 1.$
Since $a = \frac{1}{\sqrt{2}},$ this becomes:
$\frac{1}{\sqrt{2}} \sqrt{1 + \frac{b^2}{\frac{1}{2}}} = 1.$
Simplify the term inside the square root:
$1 + 2b^2.$
So we have:
$\frac{1}{\sqrt{2}} \sqrt{1 + 2b^2} = 1.$
Squaring both sides:
$\frac{1}{2} (1 + 2b^2) = 1 \implies 1 + 2b^2 = 2 \implies b^2 = \frac{1}{2}.$
Step 5: Write the equation of the hyperbola
With $a^2 = \frac{1}{2}$ and $b^2 = \frac{1}{2},$ the equation becomes:
$\frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1,$
which simplifies to
$x^2 - y^2 = \frac{1}{2}.$
Step 6: Check which point does not satisfy the hyperbola
The correct answer states that the point
$\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$
does not lie on the hyperbola. We verify:
Substitute $x = \sqrt{\frac{3}{2}}$ and $y = \frac{1}{\sqrt{2}}$ into
$x^2 - y^2 = \frac{1}{2}.$
$x^2 = \frac{3}{2}, \quad y^2 = \frac{1}{2}.$
So
$x^2 - y^2 = \frac{3}{2} - \frac{1}{2} = 1.$
However, the hyperbola’s equation demands
$x^2 - y^2 = \frac{1}{2}.$
Because $1 \neq \frac{1}{2},$ the point
$\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$
does not satisfy the hyperbola equation and hence does not lie on it.
Final Answer
The point
$\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$
does not lie on the given hyperbola.
